notes100-ch5.pdf

MATH 100 Spring 2006-07

Introduction to Multivariable Calculus

Lecture Notes

Dr. Tony Yee

Department of Mathematics

The Hong Kong University of Science and Technology

May 4, 2007

Contents

Table of Contents iii

1 Vectors and Geometry of Space 11.1 Three-Dimensional Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 Equations of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.6 Equations of Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.7 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2 Vector-Valued Functions 332.1 Vector Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2 Calculus with Vector Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.3 Tangent, Normal and Binormal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 422.4 Arc Length in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3 Partial Derivatives 493.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.5 Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.6 Applications of Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

4 Multiple Integrals 1154.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1154.2 Double Integrals Over Non-rectangular Regions . . . . . . . . . . . . . . . . . . . . . 1254.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 1364.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1484.5 Triple Integrals in Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 1554.6 Triple Integrals in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 158

5 Integration in Vector Fields 1635.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1635.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1685.3 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1905.4 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

iii

CONTENTS

5.5 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2205.6 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

iv

Chapter 5

Integration in Vector Fields

In this chapter we combine ideas from the preceding four chapters: space curves, vector functions, partialderivatives, and multiple integrals. The result is the development of line integrals, vector fields, andsurface integrals giving powerful mathematical tools for science and engineering. Line integrals are usedto find the work done by a force in moving an object along a path, and to find the mass of a wire having avarying density. Surface integrals are used to find the rate at which a fluid flows across a surface. Finally, weconclude with three major theorems, Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem.These theorems connect the new tools (i.e., the new integrals) and offer insight into their mathematicalcalculations and physical applications.

5.1 Vector Fields

We start this chapter with the definition of a vector field as they will be a major component of this chapter.Let us start with the formal definition of a vector field.

5.1.1 Definition A vector field on two- (or three-) dimensional space is a function F that assigns to eachpoint (x, y) (or (x, y, z)) a two- (or three-) dimensional vector given by F(x, y) (or F(x, y, z)).

The definition may not make a lot of sense, but most people do know what a vector field is, or at least theyhave seen a sketch of a vector field. Pick an example in physics for reference. Suppose a region in the planeor in space is occupied by a moving fluid, such as air or water. The fluid is made up of a large number ofparticles, and at any instant of time, a particle has a velocity v. At different points of position at a given(same) time, these velocities can vary. We can think of a velocity vector being attached to each point ofthe fluid representing the velocity of a particle at that point. Such a fluid flow is an example of a vectorfield. Vector fields are also associated to forces such as gravitational attraction, and to magnetic force fields,electric fields.

Generally, a vector field is a function that assigns a vector to each point in its domain. The standardnotation for a vector field F is

F(x, y) = P (x, y) i + Q(x, y) j,

F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z)k

depending on whether or not we are in two or three dimensions. The component functions P , Q, R aresometimes called scalar functions. The field is continuous if the component functions are continuous; itis differentiable if each of the component functions is differentiable.

Let us take a look at a couple of examples.

163

5. Integration in Vector Fields

¥ Example 5.1.1 (Direction fields)

Sketch each of the following direction fields.

(a) F(x, y) = −y i + x j. (b) F(x, y, z) = 2x i− 2y j− 2xk.

Solution

(a) To graph the vector field we need to get some “values” of the function. This means plugging in somepoints into the function. Here are a couple of evaluations.

F(1

2,

1

2) = −1

2i +

1

2j,

F(1

2, −1

2) =

1

2i +

1

2j,

F(3

2,

1

4) = −1

4i +

3

2j.

So, just what do these evaluations tell us? The first one tells us that at the point ( 12, 1

2) we will

plot the vector − 12i + 1

2j in the first quadrant of 2-space. Likewise, the second component tells us

that at the point ( 12, − 1

2) we will plot the vector 1

2i + 1

2j in the quadrant IV. We can continue in

this fashion plotting vectors for several points and we will get the following sketch of the vector field.Just remind that a vector must have a direction as well as a length associated to it.

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

x-axis

F(x,y)=<-y,x>

If we want significantly more points plotted then it is usually best to use a computer aided graph-ing system such as Maple or Mathematica. In fact the figures included here are generated byMathematica.

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

x-axis

F(x,y)=<-y,x>

164

5.1 Vector Fields

(b) In the case of three-dimensional vector fields it is almost always better to use Maple, Mathematica,or some other graphing tools. Despite that let us go ahead and do a couple of evaluations anyway.

F(1,−3, 2) = 2i + 6j− 2k,

F(0, 5, 3) = −10 j.

Notice that z only affect the placement of the vector in this case and does not affect the direction orthe magnitude of the vector. Sometimes this will happen so do not get excited about it when it does.The following is a sketch generated by Mathematica.

x-axis

z-axis

-5

0

5

-5

-2.5

0

2.55

-5

0

5

-5

-2.5

0

2.55

2

Now that we have seen a couple of vector fields in the previous example let us notice that in fact we havealready seen a vector field function in Chapter 2, the gradient vector of the differentiable real-valuedfunction f(x, y, z). It is the vector ∇f defined as follows:

∇f(x, y, z) =∂f

∂xi +

∂f

∂yj +

∂f

∂zk.

The partial derivatives on the right-hand side of the above equation are evaluated at the point (x, y, z).Thus ∇f(x, y, z) is a vector field. It is the gradient vector field of the function f and is sometimesdenoted by grad f . In these cases the function f(x, y, z) is often called a scalar function to differentiate itform the vector field. Also recall that the vector ∇f(x, y, z) points in the direction in which the maximaldirectional derivative of f at (x, y, z) is obtained. For example, if f(x, y, z) is the temperature at thepoint (x, y, z) in space, then you should move in the direction ∇f(x, y, z) in order to warm up the mostquickly. Besides, in the case of a two-variable scalar function f(x, y) we suppress the third component inthe previous equation, so ∇f(x, y) = fx i + fy j defines a plane vector field.

It is fairly natural to suggest the formal expression

∇ =∂

∂xi +

∂

∂yj +

∂

∂zk

165


and think of ∇ as a vector differential operator. That is, ∇ is the operation that, when applied tothe scalar function f , yields its gradient vector field ∇f . This operation behaves in several familiar andimportant ways like the operation Dx := ∂/∂x of single variable differentiation.

As a computationally useful instance, suppose that f and g are functions and that a and b areconstants. It then follows from the linearity of partial differentiation that

∇(af + bg) = a∇f + b∇g.

Thus the gradient operator is linear. It also satisfies the product rule, as shown in the following.

∇(fg) =∂(fg)

∂xi +

∂(fg)

∂yj +

∂(fg)

∂zk

= (fgx + gfx) i + (fgy + gfy) j + (fgz + gfz)k

= f · (gx i + gy j + gz k) + g · (fx i + fy j + fz k) ,

or

∇(fg) = f ∇g + g∇f .

¥ Example 5.1.2 (Gradient vector field)

Find the gradient vector field of the following functions.

(a) f(x, y) = x2 sin(5y). (b) f(x, y, z) = ze−xy.

Solution

(a) As mentioned before there is a two-dimensional definition for gradient vector field. All that we needto do is drop off the third component of the vector. Here is an example of this kind.

∇f(x, y) = 〈 ∂

∂x

`x2 sin(5y)

´,

∂

∂y

`x2 sin(5y)

´〉 = 〈2x sin(5y), 5x2 cos(5y)〉.

(b) There is not much to do here other than take the gradient.

∇f(x, y, z) = 〈−yze−xy, −xze−xy, e−xy〉.

2

Let us do another example that will illustrate the relationship between the gradient vector field of afunction and its level curves.

¥ Example 5.1.3 (Vector field and level curves)

Sketch the gradient vector field for f(x, y) = x2 + y2 as well as several level curves for this function.

166

5.1 Vector Fields

Solution Recall that the level curves for a function are nothing more than curves defined by

f(x, y) = k

for various values of k. So, for our function the level curves are defined by the equation

x2 + y2 = k

and so they are circles centered at the origin with radius√

k. The gradient vector field for this function is

∇f(x, y) = 2x i + 2y j.

Here is a sketch of several of the level curves as well as the gradient vector field.

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

2

Notice that the vectors of the vector field are all perpendicular (or orthogonal) to the level curves. This willalways be the case when we are dealing with the level curves of a function as well as its gradient vector field.

Notice as well that the closer the gradient curves are to each other the larger the vectors in the vectorfield. The closer the level curves are the faster the function is changing at that point. Also recall that thedirection of fastest change for a function is given by the gradient vector at that point. Therefore, it shouldmake sense that the two ideas should match up as they do here.

The final topic of this section is that of conservative vector fields. A vector field F is called a conser-vative vector field if there exists a function f such that F = ∇f . If F is a conservative vector fieldthen the function f is called a potential function for F.

All this definition is saying is that a vector field is conservative if it is also a gradient vector field forsome function. For instance the vector field F(x, y) = y i+x j is a conservative vector field with a potentialfunction of f(x, y) = xy because ∇f = 〈fx, fy〉 = 〈y, x〉.

On the other hand, F(x, y) = −y i + x j is not a conservative vector field since there is no function fsuch that F = ∇f . If you are not sure that you believe this at this point be patient, we will be able toprove this in a later subsection (page 185). From there we will also show how to find the potential functionfor a conservative vector field.

167


5.2 Line Integrals

In this section we are now going to introduce a new kind of integral. However, before we do that it isimportant to note that you will need to remember how to parametrize equations, or put another away, youwill need to be able to write down a set of parametric equations for a given curve. You should have seensome of this in Chapter 2. If you need some review you should go back and review some of the basics ofparametric equations and curves.

Here are some of the more basic curves that we will need to know how to do as well as limits on theparameter if they are required.

Curve Parametric Equations

x2

a2+

y2

b2= 1 (Ellipse) x = a cos t, y = b sin t, 0 6 t 6 2π

x2 + y2 = r2 (Circle) x = r cos t, y = r sin t, 0 6 t 6 2π

y = f(x) x = t, y = f(t)

x = g(y) y = t, x = g(t)

Line segment from (x0, y0, z0) to (x1, y1, z1) x = (1− t)x0 + tx1, y = (1− t) y0 + ty1,z = (1− t) z0 + tz1, 0 6 t 6 1

With the final one we just drop the z component if we need the two-dimensional version. In fact, we willbe using the two-dimensional version of this in this section.

Now let us move on to line integrals. In single variable calculus we integrate f(x), a function of a singlevariable, over an interval [a, b]. In this case we are thinking of x as taking all the values in this intervalstarting at a and ending at b. With line integrals we will start with integrating a function f(x, y), afunction of two variables, and the values of x and y that we are going to use will be the points, (x, y),that lie on a plane curve C. Note that this is different from the double integrals that we were working within the previous chapter where the points came out of some two-dimensional region.

Let us start with the two-dimensional curve C that the points come from. We will assume that thecurve is smooth (defined shortly) and is given by the parametric equations

x = x(t), y = y(t), a 6 t 6 b.

We will often want to write the parametrization of the curve as a vector function. In this case the curve is

r(t) = x(t) i + y(t) j, a 6 t 6 b.

The curve is called smooth if r′(t) is continuous and r′(t) 6= 0 for all t.

5.2.1 Definition The line integral of f(x, y) along the curve C is denoted by

Z

C

f(x, y) ds.

168

5.2 Line Integrals

We use the arc length differential ds to acknowledge the fact that we are moving along the curve, C,instead of the x-axis (denoted by dx) or the y-axis (denoted by dy). Because of the ds this is sometimescalled the line integral of f with respect to arc length.

We have seen the notation ds before. If you recall from single variable calculus when we look at thearc length of a curve given by parametric equations we find it to be

L =

Z b

a

ds, where ds =

r“dx

dt

”2

+“dy

dt

”2

dt.

It is no coincidence that we use ds for both of these problems. The ds is the same for both the arc lengthintegral and the notation for the line integral. So, to compute a line integral we will convert everythingover to the parametric equations. The line integral is then

Z

C

f(x, y) ds =

Z b

a

f“x(t), y(t)

”r“dx

dt

”2

+“dy

dt

”2

dt.

Do not forget to plug the parametric equations into the function f as well so that the integrand on theright-hand side is purely a function of t without x and y anymore. If you use the vector form of theparametrization we can simplify the notation up somewhat by noticing that

r“dx

dt

”2

+“dy

dt

”2

= ‖r′(t)‖,

where ‖r′(t)‖ is the magnitude or norm of r′(t). Using this notation the line integral becomes

Z

C

f(x, y) ds =

Z b

a

f“x(t), y(t)

”‖r′(t)‖ dt.

Note that as long as the parametrization of the curve C is traced out exactly once as t increases from ato b the value of the line integral will be independent of the parametrization of the curve.

Let us take a look at an example of a line integral.

¥ Example 5.2.1 (Line integral)

Evaluate

Z

C

xy4 ds, where C is the right half of the circle x2 + y2 = 16.

Solution We first need a parametrization of the circle. This is given by

x = 4 cos t, y = 4 sin t.

We also need a range of t’s that will give the right half of the circle. That is

−π

26 t 6 π

2.

Now, we need the derivatives of the parametric equations and let us compute ds.

dx

dt= −4 sin t,

dy

dt= 4 cos t,

ds =p

16 sin2 t + 16 cos2 t dt = 4 dt.

169


The line integral is then

Z

C

xy4 ds =

Z π/2

−π/2

4 cos t · (4 sin t)4 · 4 dt

= 4096

Z π/2

−π/2

sin4 t d(sin t)

=4096

5

hsin5 t

iπ/2

−π/2

=4096

5(1 + 1) =

8192

5.

2

Next we need to talk about line integrals over piecewise smooth curves. A piecewise smooth curve isany curve that can be written as the union of a finite number of smooth curves, C1, C2, · · · , Cn wherethe end point of Ci is the starting point of Ci+1. Below is an illustration of a piecewise smooth curve.

x

y

C1

C2

C3C4

Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do. Let C be apiecewise smooth curve. In such a case the value of a line integral along C is defined to be the sum of itsvalues along the smooth segments of C. For instance, the line integral for some function f over the abovepiecewise smooth curve C (= C1 ∪ C2 ∪ C3 ∪ C4) will be

Z

C

f(x, y) ds =

Z

C1

f(x, y) ds +

Z

C2

f(x, y) ds +

Z

C3

f(x, y) ds +

Z

C4

f(x, y) ds.

Let us see an example of this.


Evaluate

Z

C

4x3 ds, where C is the curve shown below.

170

5.2 Line Integrals

x

y

•

•C1: y = −1

C2: y = x3 − 1

C3: x = 1

−2 −1

1

2

Solution First we need to parametrize each of the curves.

C1 : x = t, y = −1, −2 6 t 6 0,

C2 : x = t, y = t3 − 1, 0 6 t 6 1,

C3 : x = 1, y = t, 0 6 t 6 2.

Now let us do the line integral over each of these curves.Z

C1

4x3 ds =

Z 0

−2

4t3p

(1)2 + (0)2 dt

=

Z 0

−2

4t3 dt =ht4i0−2

= −16,

Z

C2

4x3 ds =

Z 1

0

4t3p

(1)2 + (3t2)2 dt

=

Z 1

0

4t3p

1 + 9t4 dt =1

9

h (1 + 9t4)3/2

3/2

i10

=2

27

“103/2 − 1

”≈ 2.268,

Z

C3

4x3 ds =

Z 2

0

4(1)3p

(0)2 + (1)2 dt =

Z 2

0

4 dt = 8.

Finally, the line integral that we have asked to compute isZ

C

4x3 ds =

Z

C1

4x3 ds +

Z

C2

4x3 ds +

Z

C3

4x3 ds ≈ −16 + 2.268 + 8 = −5.732.

2

Notice that we put direction arrows on the curve in the above example. The direction of motion along acurve may change the value of the line integral as we will see in the next subsection (page 176). Also notethat the curve can be thought of a curve that takes us from the point (−2,−1) to the point (1, 2). Letus see what happens to the line integral if we change the path between these two fixed points.

171



Evaluate

Z

C

4x3 ds, where C is the line segment from (−2,−1) to (1, 2).

Solution From the parametrization formulas in the beginning of this section (page 168) we know thatthe line segment starts at (−2,−1) and ending at (1, 2) is given by

r(t) = (1− t) 〈−2,−1〉+ t 〈1, 2〉 = 〈−2 + 3t, −1 + 3t〉,for 0 6 t 6 1. This means that the individual parametric equations are

x = −2 + 3t, y = −1 + 3t.

Using this path the line integral isZ

C

4x3 ds =

Z 1

0

4(−2 + 3t)3√

9 + 9 dt

= 12√

2

Z 1

0

`27t3 − 54t2 + 36t− 8

´dt

= 12√

2

»27

4− 54

3+

36

2− 8

–= 12

√2 · (−5

4)

= −15√

2 ≈ −21.213.

2

So, the previous two examples seem to suggest that if we change the path between fixed initial and finalpoints then the value of the line integral (with respect to arc length) will change. While this will happenfairly regularly we cannot assume that it will always happen. In a later subsection (page 182) we willinvestigate this idea in more detail.

Next we will see what happens if we change the direction of a path.


Evaluate

Z

C

4x3 ds, where C is the line segment from (1, 2) to (−2,−1).

Solution This is not much different from the previous example. Here is the parametrization of the curve.

r(t) = (1− t) 〈1, 2〉+ t 〈−2,−1〉 = 〈1− 3t, 2− 3t〉,for 0 6 t 6 1. Remember that we are switching the direction of the curve and this will also change theparametrization so we can make sure that we start / end at the proper point. Here is the line integral.

Z

C

4x3 ds =

Z 1

0

4(1− 3t)3√

9 + 9 dt

= 12√

2

Z 1

0

`−27t3 + 27t2 − 9t + 1´dt

= 12√

2

»−27

4+

27

3− 9

2+ 1

–= 12

√2

„−5

4

«

= −15√

2 ≈ −21.213.

2

172

5.2 Line Integrals

So it seems when we switch the direction of the curve the line integral (with respect to arc length) willnot change. This will always be true for this kind of line integrals. However, there are other kinds of lineintegrals for which this won’t be the case. We will see more examples in the next couple of subsections(pages 176, 178).

Before working another example let us formalize this idea up somewhat. Let us suppose that the curveC has the parametrization x = x(t), y = y(t). Let us also suppose that the initial point on the curve isA and the final point on the curve is B. The parametrization x = x(t), y = y(t) will then determinean orientation for the curve where the positive direction is the direction that is traced out as t increases.Finally, let −C be the curve with the same points as C, however in this case the curve has B as theinitial point and A as the final point, again t is increasing as we traverse this curve. In other words, givena curve C, the curve −C is the same curve as C except the direction has been reversed.

We then have the following fact about line integrals with respect to arc length.

Z

C

f(x, y) ds =

Z

−C

f(x, y) ds.

So, we can change the direction of a line integral with respect to arc length and it does not change the valueof the integral. Now let us work another example.


Evaluate

Z

C

x ds for each of the following curves.

(a) C1 : y = x2, −1 6 x 6 1.

(b) C2 : The line segment from (−1, 1) to (1, 1).

(c) C3 : The line segment from (1, 1) to (−1, 1).

Solution Before working any of these line integrals let us notice that all of these curves are paths thatconnect the points (−1, 1) to (1, 1). Also notice that C3 = −C2 and so by the fact above these twoshould give the same answer. Here is a sketch of the curves C1 and C2 (C3 is skipped).

x

y

C1

C2

−1 1

1

173


(a) Here is the parametrization for this curve.

C1 : x = t, y = t2, −1 6 t 6 1.

Here is the line integral.

Z

C1

x ds =

Z 1

−1

tp

1 + 4t2 dt =1

12

h(1 + 4t2)3/2

i1−1

= 0.

(b) There are two parametrizations that we could use here for this curve. The first is to use the formulawe used in the previous couple of examples. That parametrization is

C2 : r(t) = (1− t) 〈−1, 1〉+ t 〈1, 1〉 = 〈2t− 1, 1〉,for 0 6 t 6 1. Sometimes we have no choice but to use this parametrization. However, in this casethere is a second (probably) easier parametrization. The second one uses the fact that we are reallyjust graphing a portion of the line y = 1. This parametrization is

C2 : x = t, y = 1, −1 6 t 6 1.

This will be a much easier parametrization to use so we will use this. Here is the line integral for thiscurve.

Z

C2

x ds =

Z 1

−1

t√

1 + 0 dt =h12t2i1−1

= 0.

Note that, unlike the line integral we worked with in Examples 5.2.2–5.2.3, this time we have thesame value for the integral despite the fact that the path is different. This will happen on occasion.We should also not expect this integral to be the same for all paths between these two points. Atthis moment all we know is that for these two paths the line integral will have the same value. It iscompletely possible that there is another path between these two points that will give a different valuefor the line integral.

(c) Now, according to our fact above we really don’t need to do anything here since we know thatC3 = −C2. The fact tells us that this line integral should be the same as the second part (i.e., zero).However, let us verify that, plus there is a point we need to make here about the parametrization.Here is the parametrization for this curve.

C3 : r(t) = (1− t) 〈1, 1〉+ t 〈−1, 1〉 = 〈1− 2t, 1〉,for 0 6 t 6 1. Note that this time we cannot use the second parametrization that we used in part (b)since we need to move from right to left as the parameter increases and the second parametrizationused in the previous part will move in the opposite direction. Here is the line integral.

Z

C3

x ds =

Z 1

0

(1− 2t)√

4 + 0 dt = 2ht− t2

i10

= 0.

Sure enough we have the same answer as the second part.

2

So far we have only looked at line integrals over a two-dimensional curve. However, there is no reason torestrict ourselves like that. We can do line integrals over three-dimensional curves in space as well.

174

5.2 Line Integrals

Let us suppose that the three-dimensional curve C is given by the parametrization

x = x(t), y = y(t), z = z(t), a 6 t 6 b,

then the line integral is given by

Z

C

f(x, y, z) ds =

Z b

a

f“x(t), y(t), z(t)

” r“dx

dt

”2

+“dy

dt

”2

+“dz

dt

”2

dt.

Note that often when dealing with three-dimensional space the parametrization will be given as a vectorfunction.

r(t) = 〈x(t), y(t), z(t)〉.Notice that we changed up the notation for the parametrization a little. Since we rarely use the functionnames we simply kept the x, y, and z and added on the (t) part to denote that they are functions of theparameter. Also notice that, as with two-dimensional curves, we have

r“dx

dt

”2

+“dy

dt

”2

+“dz

dt

”2

= ‖r′(t)‖,

and the line integral can again be written as

Z

C

f(x, y, z) ds =

Z b

a

f“x(t), y(t), z(t)

”‖r′(t)‖ dt.

So, outside of the addition of a third parametric equation line integrals in three-dimensional space workexactly the same as those in two-dimensional space. Let us work a quick example.


Evaluate

Z

C

xyz ds, where C is the circular helix given by r(t) = 〈cos t, sin t, 3t〉, 0 6 t 6 4π.

Solution Note that we first saw the vector equation for a helix back in Chapter 2 where you can find aquick sketch of a similar helix (page 36). Here is the line integral.

Z

C

xyz ds =

Z 4π

0

3t cos t sin tp

sin2 t + cos2 t + 9 dt

=

Z 4π

0

3t · 1

2sin 2t

√1 + 9 dt

=3√

10

2

Z 4π

0

t sin 2t dt

=3√

10

2

»1

4sin 2t− t

2cos 2t

–4π

0

= −3√

10 π.

Remark that here we have used the integration technique called integration by parts. So, as we can seethere really is not too much difference between two- and three-dimensional line integrals. 2

175


• Line integrals with respect to coordinate variables

Previously we looked at line integrals with respect to arc length. Now we want to look at line integrals withrespect to x, and/or y. We will start with a two-dimensional curve C with parametrization

x = x(t), y = y(t), a 6 t 6 b.

5.2.2 Definition The line integral of f(x, y) with respect to x is

Z

C

f(x, y) dx =

Z b

a

f“x(t), y(t)

”x′(t) dt,

while the line integral of f(x, y) with respect to y is

Z

C

f(x, y) dy =

Z b

a

f“x(t), y(t)

”y′(t) dt.

Note that the only notational difference between these two and the line integral with respect to arc lengthis the differential. These have a dx or dy while the line integral with respect to arc length has a ds. Sowhen evaluating line integrals be careful to first see which differential you are having so you don’t work thewrong kind of line integral.

These two integrals often appear together and so we have the shorthand notation for these cases.

Z

C

P dx + Q dy =

Z

C

P (x, y) dx +

Z

C

Q(x, y) dy.

Although it would be natural enough to writeRC

(P dx + Q dy) on the left-hand side in the above formula,

the parentheses are customarily omitted. Let us take a look at an example of this kind of line integral.


Evaluate

Z

C

sin πy dy + yx2 dx, where C is the line segment from (0, 2) to (1, 4).

Solution Here is the parametrization of the curve.

r(t) = (1− t) 〈0, 2〉+ t 〈1, 4〉 = 〈t, 2 + 2t〉, 0 6 t 6 1.

The line integral isZ

C

sin πy dy + yx2 dx =

Z 1

0

sin(π(2 + 2t)) · 2 dt +

Z 1

0

(2 + 2t)(t)2 · 1 dt

=1

π

hcos(2π + 2πt)

i10

+h23t3 +

1

2t4i10

=7

6.

2

Previously we saw that changing the direction of the curve for a line integral with respect to arc length doesnot change the value of the integral. Let us see what happens with line integrals with respect to x and/or y.

176

5.2 Line Integrals


Evaluate

Z

C

sin πy dy + yx2 dx, where C is the line segment from (1, 4) to (0, 2).

Solution So, we simply change the direction of the curve. Here is the new parametrization.

r(t) = (1− t) 〈1, 4〉+ t 〈0, 2〉 = 〈1− t, 4− 2t〉, 0 6 t 6 1.

The line integral isZ

C

sin πy dy + yx2 dx =

Z

C

sin πy dy +

Z

C

yx2 dx

=

Z 1

0

sin(π(4− 2t)) · (−2) dt +

Z 1

0

(4− 2t)(1− t)2 · (−1) dt

=1

π

hcos(4π − 2πt)

i10−h− 1

2t4 +

8

3t3 − 5t2 + 4t

i10

= −7

6.

2

So, switching the direction of the curve gives us a different value or at least the opposite sign of the valuefrom the first example. In fact this will always happen with this kind of line integrals (with respect to xand / or y). If C is any curve, then

Z

−C

f(x, y) dx = −Z

C

f(x, y) dx and

Z

−C

f(x, y) dy = −Z

C

f(x, y) dy.

With the combined form of these two integrals, we get

Z

−C

P dx + Q dy = −Z

C

P dx + Q dy.

We can also do these integrals over three-dimensional curves as well. In this case we will pick up a thirdintegral (with respect to z) and the three integrals will be

Z

C

f(x, y, z) dx =

Z b

a

f“x(t), y(t), z(t)

”x′(t) dt,

Z

C

f(x, y, z) dy =

Z b

a

f“x(t), y(t), z(t)

”y′(t) dt,

Z

C

f(x, y, z) dz =

Z b

a

f“x(t), y(t), z(t)

”z′(t) dt,

where the curve C is parametrized by x = x(t), y = y(t), z = z(t), a 6 t 6 b. As with thetwo-dimensional version these three will often occur together so the shorthand we will be using here is

Z

C

P dx + Q dy + R dz =

Z

C

P (x, y, z) dx +

Z

C

Q(x, y, z) dy +

Z

C

R(x, y, z) dz.

Let us work an example.

177



Evaluate

Z

C

y dx + x dy + z dz, where C is given by x = cos t, y = sin t, z = t2, 0 6 t 6 2π.

Solution We have the curve parametrized so there is not much to do other than evaluate the integral.Z

C

y dx + x dy + z dz =

Z

C

y dx +

Z

C

x dy +

Z

C

z dz

=

Z 2π

0

sin t(− sin t) dt +

Z 2π

0

cos t(cos t) dt +

Z 2π

0

t2(2t) dt

= −Z 2π

0

sin2 t dt +

Z 2π

0

cos2 t dt +

Z 2π

0

2t3 dt

= −1

2

Z 2π

0

(1− cos 2t) dt +1

2

Z 2π

0

(1 + cos 2t) dt +

Z 2π

0

2t3 dt

=

»−1

2

„t− 1

2sin 2t

«+

1

2

„t +

1

2sin 2t

«+

1

2t4–2π

0

=

»−1

2sin 2t +

1

2t4–2π

0

= 8π4.

2

• Line integrals of vector fields

So far we have looked at two kinds of line integrals of functions (with respect to arc length as well ascoordinate variables). In this subsection we are going to evaluate line integrals of vector fields (Section 5.1).We will start with the vector field


and the three-dimensional, smooth curve is given by

r(t) = x(t) i + y(t) j + z(t)k, a 6 t 6 b.

5.2.3 Definition The line integral of the vector field F along C is

Z

C

F · dr =

Z b

a

F“r(t)

” · r′(t) dt.

Note the notation in the left side. That really is a dot product of the vector field and the differential and

the differential dr really is a vector. Also, F“r(t)

”is a shorthand for

F“r(t)

”= F

“x(t), y(t), z(t)

”.

178

5.2 Line Integrals

We can also write the line integral of vector field as a line integral with respect to arc length.

Z

C

F · dr =

Z

C

F · T ds,

where T(t) is the unit tangent vector to the curve C and is given by

T =r′(t)‖r′(t)‖ , where

r′(t) =dx

dti +

dy

dtj +

dz

dtk,

‖r′(t)‖ =

r“dx

dt

”2

+“dy

dt

”2

+“dz

dt

”2„

=ds

dt

«.

In words, the integral of a vector field along a curve has the same value as the integral of the tangentialcomponent of the vector field along the curve. In the following we may give the detail proof of

T ds = dr,

which is

T ds =r′(t)‖r′(t)‖ · ‖r

′(t)‖ dt

= r′(t) dt

=

„dx

dti +

dy

dtj +

dz

dtk

«dt

= dx i + dy j + dz k = dr.

Let us give one physical interpretation of the line integral. Suppose that the vector field F representsa force throughout a region in space (it might be the force of gravity or an electromagnetic force of somekind) and that r(t) = x(t) i + y(t) j + z(t)k is a smooth curve in the region. Then the work done by theforce F in moving an object over the smooth curve from t = a to t = b is given by

Work done = W =

Z

C

F · T ds.

Thus work is the integral with respect to arc length of the tangential component of the force. Intuitively, wemay regard dW = F · T ds as the infinitesimal element of work done by the tangential component F · Tof the force in moving the object along the arc length element ds. The line integral is then the “sum” ofall these infinitesimal elements of work.

In general we use the first form

Z

C

F · dr to compute these line integrals as it is usually much easier

to use. Let us take a look at a couple of examples.

179



EvaluateZ

C

F · dr,

where F(x, y, z) = 8x2yz i + 5z j− 4xy k and C is the curve given by r(t) = t i + t2 j + t3 k, 0 6 t 6 1.

Solution We first need the vector field evaluated along the curve.

F“r(t)

”= 8t2(t2)(t3) i + 5t3 j− 4t(t2)k = 8t7 i + 5t3 j− 4t3 k.

Next we need the derivative of the parametrization.

r′(t) = i + 2t j + 3t2 k.

Finally, let us get the dot product

F“r(t)

” · r′(t) = 8t7 + 10t4 − 12t5.

The line integral is then

Z

C

F · dr =

Z 1

0

`8t7 + 10t4 − 12t5

´dt =

ht8 + 2t5 − 2t6

i10

= 1.

2


EvaluateZ

C

F · dr,

where F(x, y, z) = xz i− yz k and C is the line segment from (−1, 2, 0) to (3, 0, 1).

Solution We will first need the parametrization of the line segment. Here is the parametrization.

r(t) = (1− t) 〈−1, 2, 0〉+ t 〈3, 0, 1〉 = 〈4t− 1, 2− 2t, t〉, 0 6 t 6 1.

So, let us get the vector field evaluated along the curve.

F“r(t)

”= (4t− 1)(t) i− (2− 2t)(t)k = (4t2 − t) i− (2t− 2t2)k.

Now we need the derivative of the parametrization.

r′(t) = 〈4,−2, 1〉.The dot product is

F“r(t)

” · r′(t) = 4(4t2 − t)− (2t− 2t2) = 18t2 − 6t.

The line integral becomes

Z

C

F · dr =

Z 1

0

`18t2 − 6t

´dt =

h6t3 − 3t2

i10

= 3.

2

180

5.2 Line Integrals

Let us close this subsection out by generally getting a nice relationship between line integrals of vector fieldsand line integrals with respect to coordinate variables x, y, and z.

Given the vector field


and the curve C parametrized by

r(t) = x(t) i + y(t) j + z(t)k, a 6 t 6 b,

the line integral isZ

C

F · dr =

Z b

a

(P i + Q j + Rk) · `x′ i + y′ j + z′ k´

dt

=

Z b

a

P x′ + Q y′ + R z′ dt

=

Z b

a

P x′ dt +

Z b

a

Q y′ dt +

Z b

a

R z′ dt

=

Z

C

P dx +

Z

C

Q dy +

Z

C

R dz

=

Z

C

P dx + Q dy + R dz.

So we see thatZ

C

F · dr =

Z

C

P dx + Q dy + R dz.

Note that this gives us another method for evaluating line integrals of vector fields. This also allows us tosay the following about reversing the direction of the path with line integrals of vector fields.

Z

−C

F · dr = −Z

C

F · dr.

This should make enough sense given that we know that this is true for line integrals with respect tocoordinate variable x, y, and / or z and the line integrals of vector fields can be defined in terms of lineintegrals with respect to x, y, and z.

¥ Example 5.2.12 (Work as a line integral)

The work done by the force field F = y i + z j + xk in moving a particle from (0, 0, 0) to (1, 1, 1) alongthe twisted cubic x = t, y = t2, z = t3 is given by the line integral

W =

Z

C

F · dr =

Z

C

F · T ds =

Z

C

y dx + z dy + x dz,

and we can easily compute the value of the line integral by either one of the above forms. The last one

would be more appropriate and is equal to

Z 1

0

ˆt2(1) + t3(2t) + t(3t2)

˜dt, or the value is W = 89

60. 2

181


• Fundamental theorem for line integrals

In single variable calculus we have the Fundamental Theorem of Calculus that tells us how to evaluatedefinite integrals. This means

Z b

a

f ′(x) dx = f(b)− f(a).

It turns out that there is a version of this for line integrals over certain kinds of vector fields. Here it is.

Theorem 5.2.1 Suppose that C is a smooth curve given by r(t), a 6 t 6 b. Also suppose that fis a function whose gradient vector, ∇f , is continuous on C. Then

Z

C

∇f · dr = f“r(b)

”− f

“r(a)

”.

Note that r(a) represents the initial point on C while r(b) represents the final point on C. Also, wedid not specify the number of variables for the function f since it is really immaterial to the theorem. Thetheorem will hold regardless of the number of variables in the function.

Let us take a quick look at an example of using this theorem when f is a function of three variables.


EvaluateZ

C

∇f · dr,

where f(x, y, z) = cos πx+sin πy−xyz and C is any path that starts at (1,1

2, 2) and ends at (2, 1,−1).

Solution First let us notice that we did not specify the path for getting from the first point to the secondpoint. The reason for this is simple. The theorem above tells us that all we need are the initial and finalpoints on the curve in order to evaluate this kind of line integral. So, let r(t), a 6 t 6 b be any path thatstarts at (1, 1

2, 2) and ends at (2, 1,−1). Then

r(a) = 〈1,1

2, 2〉, r(b) = 〈2, 1,−1〉.

The integral is thenZ

C

∇f · dr = f(2, 1,−1)− f(1,1

2, 2)

= cos 2π + sin π − 2(1)(−1)−„

cos π + sinπ

2− 1(

1

2)(2)

«= 4.

Notice that we also did not need the gradient vector to actually do this line integral. However, for thepractice of finding gradient vectors here it is

∇f = 〈−π sin(πx)− yz, π cos(πy)− xz, −xy〉.2

182

5.2 Line Integrals

The most important idea to get from this example is not how to compute the integral as that’s often prettysimple, all that we do is plug the final point and initial point into the function and subtract the two functionvalues. The important idea from this example (and hence about the fundamental theorem for line integrals)is that we did not really need to know the path to get the answer. In other words, we could use any pathwe want and we will always get the same result.

In the previous part on line integrals (even though we were not looking at vector fields) we saw thatoften when we change the path we will change the value of the line integral (see Examples 5.2.2–5.2.3,page 170). We now have a type of line integral for which we know that changing the path will NOT changethe value of the line integral. This is a new topic called the independence of path.

Of course, not every line integral of the formRC

F · dr is independent of path just as not every definite

integralR b

af(x) dx can be evaluated by Fundamental Theorem of Calculus. For an example, f(x) = sin x

x

has no primitive function.

Let us formalize this idea up a little. Here are some definitions. The first one we have already seenbefore, but it is important here so we will list it again. The remaining definitions are new.

5.2.4 Definition Suppose that F is a continuous vector field in some domain R.

1. F is a conservative vector field if there is a function f such that F = ∇f . The function fis called a potential function for the vector field. We first saw this definition in Section 5.1(page 167).

2.

Z

C

F · dr is independent of path if

Z

C1

F · dr =

Z

C2

F · dr for any two paths C1 and C2 in

R with the same initial and final points.

3. A path C is called closed if its initial and final points are the same point. For example a circleis a closed path.

4. A path C is simple if it does not cross itself. A circle is a simple curve while a curve withshape “ ∞” is not simple.

5. A region R is open if it does not contain any of its boundary points.

6. A region R is connected if we can connect any two points in the region with a path that liescompletely in R.

7. A region R is simply-connected if it is connected and it contains no holes. We will not needthis one until the next subsection (page 185), but it fits in with all the other definitions given hereso this is a natural place to put the definition.

With these definitions we can now give some nice facts. These are some nice facts to remember as we workwith line integrals over vector fields. Also notice that Facts 2 & 3 and 4 & 5 are converses of each other.

183


Fact 4

1.

Z

C

∇f · dr is independent of path.

This is easy enough to justify since all we need to do is look at the theorem above. The theorem tellsus that in order to evaluate this integral all we need are the initial and final points of the curve.This in turn tells us that the line integral must be independent of path.

2. If F is a conservative vector field then

Z

C


This fact is also easy enough to justify. If F is conservative then there exists a potential function

f and so the line integral become

Z

C

F · dr =

Z

C

∇f · dr. Then using the first fact we know that

this line integral must be independent of path.

3. If F is a continuous vector field on an open connected region R and if

Z

C

F · dr is independent

of path (for any path in R) then F is a conservative vector field on R.

4. If

Z

C

F · dr is independent of path then

Z

C

F · dr = 0 for every closed path C.

5. If

Z

C

F · dr = 0 for every closed path C then

Z

C

F · dr is independent of path.

The word conservative comes from Physics, where it refers to fields in which the principle of conservationof energy holds. In gravitational and electric fields, the amount of work it takes to move a mass or chargefrom one point to another depends on the initial and final positions of the object – not on which path istaken between these positions. The fields with this property are called conservative and the line integralRC

F · dr represents the work done in moving the object.

In terms of mathematics, we come up with two major methods (up to this moment it is two, we will seethe third one in Section 5.3 called Green’s Theorem) for evaluating line integrals. They are

(1) Direct Evaluation by Parametrization of Curve C. (Examples 5.2.1–5.2.11)

(2) Fundamental Theorem for Line Integrals. (Examples 5.2.13, 5.2.17)

The first one will always work as long as you are able to find a parametrization of C but the computationsinvolved could be long and troublesome. The second one often requires less and simpler calculations but itdoes have a limitation that the vector field F must be conservative.

184

5.2 Line Integrals

• Conservative vector fields

By the previous subsection (fundamental theorem for line integrals) we know that if the vector field Fis conservative then

RC

F · dr is independent of path. This in turn means that we can easily evaluate this

line integral provided that we can find a potential function for F. In this subsection we want to look attwo questions. First, given a vector field F is there any way of determining if it is a conservative vectorfield or not? Secondly, if we know that F is indeed a conservative vector field how do we find a potentialfunction for the vector field?

The first question is easy to answer at this moment if we have two-dimensional vector field. For three-dimensional vector fields we will need to wait until the time when we discuss curl and divergence (page 197)

to answer this question. With that being said let us see how we do it for two-dimensional vector fields.

Theorem 5.2.2 (Conservative vector field) Let F = P i + Q j be a vector field on an open andsimply-connected region D. Then if P and Q have continuous first-order partial derivatives in D and

∂Q

∂x=

∂P

∂y

the vector field F is conservative.

¥ Example 5.2.14 (Conservative vector fields)

Determine if the following vector fields are conservative or not.

(a) F(x, y) =`x2 − yx

´i +`y2 − xy

´j. (b) F(x, y) =

`2xexy + x2yexy

´i +`x3exy + 2y

´j.

Solution There really is not too much to do here. All we need to do is identify P and Q then take acouple of derivatives and compare the results.

(a) In this case here is P and Q and the corresponding partial derivatives are

P = x2 − yx,∂P

∂y= −x,

Q = y2 − xy,∂Q

∂x= −y.

So, since the two partial derivatives are not the same this vector field is NOT conservative.

(b) Here is P and Q as well as the partial derivatives are

P = 2xexy + x2yexy,∂P

∂y= 2x2exy + x2exy + x3yexy = 3x2exy + x3yexy,

Q = x3exy + 2y,∂Q

∂x= 3x2exy + x3yexy.

The two partial derivatives are equal and so this is a conservative vector field. 2

185


Now that we know how to identify if a two-dimensional vector field is conservative we need to address howto find a potential function for the vector field. This is actually a fairly simple process. First, let us assumethat the vector field is conservative and so we know that a potential function, f(x, y) exists. We can thensay that

∇f =∂f

∂xi +

∂f

∂yj = P i + Q j = F.

Or by setting components equal we have

∂f

∂x= P and

∂f

∂y= Q.

By integrating each of these with respect to the appropriate variable we can arrive at the two equations

f(x, y) =

ZP (x, y) dx and f(x, y) =

ZQ(x, y) dy.

We saw this kind of integral briefly at the end of the section on iterated integrals in Chapter 4 (page 124).

It is usually best to see how we use these two facts to find a potential function in the following twoexamples.

¥ Example 5.2.15 (Conservative vector fields)

Determine if the following vector fields are conservative and find a potential function for the vector field ifit is conservative.

(a) F(x, y) =`2x3y4 + x

´i +`2x4y3 + y

´j. (b) F(x, y) =

`2xexy + x2yexy

´i +`x3exy + 2y

´j.

Solution

(a) Let us first identify P and Q and then check that the vector field is conservative or not.

P = 2x3y4 + x,∂P

∂y= 8x3y3,

Q = 2x4y3 + y,∂Q

∂x= 8x3y3.

So, the vector field F is conservative. Now let us find the potential function for F. From the firstfact above we know that

∂f

∂x= 2x3y4 + x,

∂f

∂y= 2x4y3 + y.

From these we can see that

f(x, y) =

Z `2x3y4 + x

´dx f(x, y) =

Z `2x4y3 + y

´dy.

We can use either of these to start the process of hunting f(x, y). Recall that we have mentioned(page 124) that we have to be careful with the “constant of integration” which ever integral we chooseto use. For this example let us work with the first integral and that means we are asking whatfunction do we differentiate with respect to x to get the integrand. This means that the “constant

186

5.2 Line Integrals

of integration” is going to be a function of y and/or constants will differentiate to zero when takingthe partial derivative with respect to x. Here is the first integral.

f(x, y) =

Z `2x3y4 + x

´dx =

1

2x4y4 +

1

2x2 + h(y),

where h(y) is the “constant of integration”. We now need to determine h(y). This is easier thatit might at first appear to be. To get to this point we have used the fact that we know P , but wewill also need to use the fact that we know Q to complete the problem. Recall that Q is really thederivative of f with respect to y. So, if we differentiate our function with respect to y we knowwhat it should be. So, let us differentiate f (including the h(y)) with respect to y and set it equalto Q since that is what the derivative is supposed to be.

∂f

∂y= 2x4y3 + h′(y) = 2x4y3 + y = Q.

From this we can see that

h′(y) = y.

Notice that since h′(y) is a function of y so if there are any x’s in the equation at this point wewill know that we have made a mistake. At this point finding h(y) is simple.

h(y) =

Zh′(y) dy =

Zy dy =

1

2y2 + C.

So, putting this all together we can see that a potential function for the vector field is

f(x, y) =1

2x4y4 +

1

2x2 +

1

2y2 + C.

Note that we can always check our work by verifying that ∇f = F. Also note that because theconstant C can be anything there are an infinite number of possible potential functions, althoughthey will only vary by an additive constant.

(b) Using exactly the same approach as in part (a) we might go a lot faster for this case without goingthrough much explanation. In Example 5.2.14 we have already verified that this vector field isconservative. Let us start with the following.

∂f

∂x= 2xexy + x2yexy,

∂f

∂y= x3exy + 2y.

This means that we can do either of the following integrals

f(x, y) =

Z `2xexy + x2yexy´ dx or f(x, y) =

Z `x3exy + 2y

´dy.

While we can do either of these the first integral would be somewhat more complicated as we wouldneed to do integration by parts on each term. But then the second integral is fairly simple since thesecond term only involves y’s and the first term can be done easily since x is treated as a constant.So, from the second integral we get

f(x, y) = x2exy + y2 + h(x).

Notice that this time the “constant of integration” will be a function of x. If we differentiate this withrespect to x and set equal to P we get

∂f

∂x= 2xexy + x2yexy + h′(x) = 2xexy + x2yexy = P.

187


By comparing both sides,

h′(x) = 0 =⇒ h(x) = C.

So, in this case the “constant of integration” is really a constant. Sometimes this will happen andsometimes it won’t. Here is the potential function for this vector field.

f(x, y) = x2exy + y2 + C.

2

Now, as noted above we don’t have a way (yet) of determining if a three-dimensional vector field is conser-vative or not. However, if we are given that a three-dimensional vector field is conservative find a potentialfunction is similar to the above process, although the work will be a little more involved. In this case wewill use the fact that

∇f =∂f

∂xi +

∂f

∂yj +

∂f

∂zk = P i + Q j + Rk = F.

Let us take a quick look at an example.

¥ Example 5.2.16 (Potential function)

Find a potential function for the vector field

F = 2xy3z4 i + 3x2y2z4 j + 4x2y3z3 k.

Solution We will begin with the following equalities.

∂f

∂x= 2xy3z4,

∂f

∂y= 3x2y2z4,

∂f

∂z= 4x2y3z3.

To get started we can integrate the first one with respect to x, the second one with respect to y, or the thirdone with respect to z. Let us integrate the first one with respect to x.

f(x, y, z) =

Z2xy3z4 dx = x2y3z4 + g(y, z).

Note that this time the “constant of integration” will be a function of both y and z since differentiatinganything of that form with respect to x will differentiate to zero.

Now, we can differentiate this with respect to y and set it equal to Q. Doing this gives

∂f

∂y= 3x2y2z4 + gy(y, z) = 3x2y2z4 = Q.

Of course we will need to take the partial derivative of the constant of integration since it is a function oftwo variables. We now have

gy(y, z) = 0 =⇒ g(y, z) = h(z).

Since differentiating g(y, z) with respect to y gives zero then g(y, z) could at most be a function of z.This means that we now know the potential function must be in the following form

f(x, y, z) = x2y3z4 + h(z).

To finish this out all we need to do is differentiate it with respect to z and set the result equal to R.

∂f

∂z= 4x2y3z3 + h′(z) = 4x2y3z3 = R.

188

5.2 Line Integrals

So,

h′(z) = 0 =⇒ h(z) = C.

The potential function for this vector field is then

f(x, y, z) = x2y3z4 + C.

2

Note that to keep the work to a minimum we used a fairly simple potential function for this example. It issometimes possible to guess what the potential function is based simply on the vector field. However, weshould be careful to remember that this usually won’t be the case and often this hunting process is required.

Let us recall that if F is a conservative vector field then there is a function f such that F = ∇f and

Z

C

F · dr =

Z

C


Combining the above with Theorem 5.2.1 (page 182) provides a way of evaluating the line integral

Z

C

F · dr.We need to work one final example in this section.

¥ Example 5.2.17 (Potential function)

Evaluate

Z

C

F · dr where F =`2x3y4 + x

´i +`2x4y3 + y

´j and C is given by

r(t) = (t cos(πt)− 1) i + sinπt

2j, 0 6 t 6 1.

Solution Let us take advantage of the fact (Example 5.2.15, page 186) that we know this vector field isconservative and that a potential function for the vector field is

f(x, y) =1

2x4y4 +

1

2x2 +

1

2y2 + C.

Using this we know that the integral must be independent of path and so all we need to do is just useTheorem 5.2.1 to do the evaluation.

Z

C

F · dr =

Z

C

∇f · dr = f“r(1)

”− f

“r(0)

”,

where

r(1) = 〈−2, 1〉, r(0) = 〈−1, 0〉.

So, the integral is

Z

C

F · dr = f(−2, 1)− f(−1, 0) = (8 + 2 +1

2+ C)− (

1

2+ C) = 10.

2

189


5.3 Green’s Theorem

In this section we are going to investigate the relationship between line integrals with respect to coordinatevariables (on closed paths) and double integrals. There is a linkage built into them.

Let us start off with a simple closed curve C and let R be the region enclosed by the curve. Here isa sketch of such a curve and region.

Curve C

EnclosedRegion R

First, notice that because the curve is simple and closed (page 183) there are no holes in the region R.Also notice that a direction has been put on the curve. We will use the convention here that the curve Chas a positive orientation if it is traced out in a counter-clockwise direction. Another way to think ofa positive orientation (that will cover much more general curves as we will see later) is that as we move onthe path following the positive orientation the region R is always be on the left side.

Given curves / regions such as this we have the following theorem.

Theorem 5.3.1 (Green’s theorem) Let C be a positively oriented, piecewise smooth, simple, closedcurve and let R be the region enclosed by the curve. If P and Q have continuous first-order partialderivatives on R then

Z

C

P dx + Q dy =

ZZ

R

„∂Q

∂x− ∂P

∂y

«dA.

Before working some examples there are some alternate notations that we need to acknowledge. Whenworking with a line integral in which the path satisfies the condition of Green’s Theorem we will oftendenote the line integral as

I

C

P dx + Q dy.

With the above notation it is assumed that C does satisfy the conditions of Green’s Theorem so becareful in using them. Also, sometimes the curve C is not thought of as a separate curve but instead asthe boundary curve of some region R. Let us work a couple of examples.

190


¥ Example 5.3.1 (Green’s theorem)

Use Green’s Theorem to evaluateI

C

xy dx + x2y3 dy

where C is the triangle with vertices (0, 0), (1, 0), (1, 2).

Solution Let us first sketch C and R for this case to make sure that the conditions of Green’s Theoremare satisfied for C and will need the sketch of R to evaluate the double integral.

y = 2x

y

x0

Closed Curve C

EnclosedRegion R

1

So, the curve does satisfy the conditions of Green’s Theorem and we can see that the following inequalitieswill define the region enclosed.

0 6 x 6 1, 0 6 y 6 2x.

We can identify P and Q from the line integral. They are

P = xy, Q = x2y3.

Using Green’s Theorem the line integral becomesI

C

xy dx + x2y3 dy =

ZZ

R

`2xy3 − x

´dA

=

Z 1

0

Z 2x

0

`2xy3 − x

´dy dx

=

Z 1

0

»1

2xy4 − xy

–2x

0

dx

=

Z 1

0

`8x5 − 2x2´ dx

=

»4

3x6 − 2

3x3

–1

0

=2

3.

2

191



EvaluateI

C

y3 dx− x3 dy,

where C is the positively oriented circle of radius 2 centered at the origin.

Solution A circle will satisfy the conditions of Green’s Theorem since it is closed and simple and so wedon’t have to sketch it here. Let us first identify P and Q from the line integral.

P = y3, Q = −x3.

Be careful with the minus sign on Q. Using Green’s Theorem the line integral becomes

I

C

y3 dx− x3 dy =

ZZ

R

`−3x2 − 3y2´ dA,

where R is the disk of radius 2 centered at the origin. So, since R is a disk it seems like the best way todo this integral is to use polar coordinates. Here is the evaluation of the integral.

I

C

y3 dx− x3 dy = −3

ZZ

R

`x2 + y2´ dA

= −3

Z 2π

0

Z 2

0

r3 dr dθ

= −3

Z 2π

0

»1

4r4

–2

0

dθ

= −3

Z 2π

0

4 dθ = −24π.

2

So, Green’s theorem, as stated, will not work on regions that have holes in them. However, many regionsdo have holes in them. So, let us see how we can deal with those kinds of regions.

Let us start with the following region. Even though this region does not have any holes in it the argumentsthat we are going to go through will be similar to those that we would need for regions with holes in them,except it will be a little easier to deal with and write down.

C1

−C3 C3

C2

R2R1

The region R will be R1∪R2 and recall that the symbol ∪ is called the union and means that R consistsof both R1 and R2. The boundary of R1 is C1 ∪ C3 while the boundary of R2 is C2 ∪ (−C3) and

192


notice that both of these boundaries are positively oriented. As we traverse each boundary the correspondingregion is always on the left side. Finally, also note that we can think of the whole boundary C as

C = (C1 ∪ C3) ∪ (C2 ∪ (−C3)) = C1 ∪ C2

since both C3 and −C3 will “cancel” each other out.

Now, let us start with the following double integral and use a basic property of double integrals to breakit up.

ZZ

R

(Qx − Py) dA =

ZZ

R1∪R2

(Qx − Py) dA =

ZZ

R1

(Qx − Py) dA +

ZZ

R2

(Qx − Py) dA.

Next, use Green’s Theorem on each of these and again use the fact that we can break up line integrals intoseparate line integrals for each portion of the boundary.

ZZ

R

(Qx − Py) dA =

ZZ

R1

(Qx − Py) dA +

ZZ

R2

(Qx − Py) dA

=

I

C1∪C3

P dx + Q dy +

I

C2∪(−C3)

P dx + Q dy

=

I

C1

P dx + Q dy +

I

C3

P dx + Q dy +

I

C2

P dx + Q dy +

I

−C3

P dx + Q dy.

Next, we will use the fact thatI

−C3

P dx + Q dy = −I

C3

P dx + Q dy.

Recall that changing the orientation of a curve with line integrals with respect to coordinate variables xand/or y will simply change the sign on the integral. Using this fact we get

ZZ

R

(Qx − Py) dA =

I

C1

P dx + Q dy +

I

C3

P dx + Q dy +

I

C2

P dx + Q dy −I

C3

P dx + Q dy

=

I

C1

P dx + Q dy +

I

C2

P dx + Q dy.

Finally, put the line integrals back together and we getZZ

R

(Qx − Py) dA =

I

C1

P dx + Q dy +

I

C2

P dx + Q dy

=

I

C1∪C2

P dx + Q dy

=

I

C

P dx + Q dy.

So, what did we learn from this? If you think about it this was just a lot of work and all we got out of itwas the result from Green’s Theorem which we already knew to be true. What this exercise has shownus is that if we break a region up as we did above then the portion of the line integral on the pieces of thecurve that are in the middle of the region (each of which are in the opposite direction) will cancel out. Thisidea will help us in dealing with regions that have holes in them.

193


To see this let us look at a ring.

C1

C2

R

Notice that both of the curves are oriented positively since the region R is on the left side as we traversethe curve in the indicated direction. Note as well that the curve C2 seems to violate the original definitionof positive orientation. We originally say (page 190) that a curve has a positive orientation if it is traversedin a counter-clockwise direction. However, this is only for regions that do not have holes. For the boundaryof the hole this definition won’t work and we need to resort to the second definition that we gave above.

Now, since this region has a hole in it we will apparently not be able to use Green’s Theorem on anyline integral with the curve C = C1 ∪ C2. However, if we cut the disk in half and rename all the variousportions of the curves we get the following sketch.

C1

C2

C3

C5

−C5

R1

R2

C6

−C6

C4

The boundary of the upper portion R1 of the disk is C1 ∪ C2 ∪ C5 ∪ C6 and the boundary on the lowerportion R2 is C3 ∪ C4 ∪ (−C5) ∪ (−C6). Also notice that we can use Green’s Theorem on each of thesenew sub-regions since they do not have any holes in them. This means that we can do the following.

ZZ

R

(Qx − Py) dA =

ZZ

R1

(Qx − Py) dA +

ZZ

R2

(Qx − Py) dA

=

I

C1∪C2∪C5∪C6

P dx + Q dy +

I

C3∪C4∪(−C5)∪(−C6)

P dx + Q dy.

194


Now, we can break up the line integral into line integrals on each piece of the boundary. Also recall from thework above that boundaries that have the same curve, but opposite direction will cancel. Doing this gives

ZZ

R

(Qx − Py) dA =

ZZ

R1

(Qx − Py) dA +

ZZ

R2

(Qx − Py) dA

=

I

C1

P dx + Q dy +

I

C2

P dx + Q dy +

I

C3

P dx + Q dy +

I

C4

P dx + Q dy.

But at this point we can add the line integrals back up as follows.ZZ

R

(Qx − Py) dA =

I

C1∪C2∪C3∪C4

P dx + Q dy =

I

C

P dx + Q dy.

The end result of all of these is that we could have just used Green’s Theorem on the disk from thebeginning even though there is a hole in it. This will be true in general for regions that have holes in them.

Let us take a look at an example.


EvaluateI

C

y3 dx− x3 dy,

where C consists of the circles of radius 2 and radius 1 centered at the origin.

Solution Notice that this is the same line integral as we looked at in Example 5.3.2 (page 192) and onlythe curve C is changed. In this case the region R will now be the region between these two circles andthat will only change the limits in the double integral so we will not put in some of the details here. Thefollowing is the work for this integral.

I

C

y3 dx− x3 dy = −3

ZZ

R

`x2 + y2´ dA

= −3

Z 2π

0

Z 2

1

r3 dr dθ = −3

Z 2π

0

»1

4r4

–2

1

dθ

= −3

Z 2π

0

15

4dθ = −45π

4.

2

We will finish this part with an interesting application of Green’s Theorem. Recall that we can determinethe area of a region R with the following double integral.

Area =

ZZ

R

1 dA.

Compare this double integral with the one appearing in Green’s Theorem (page 190). Here the integrandis identically one.

195


Let us think of this double integral as the result of using Green’s Theorem. In other words, let us assumethat the integrand is 1, i.e.,

Qx − Py = 1

and see if we can get some functions P and Q that will satisfy this.

There are many functions that will satisfy this. Here are some of the more common functions.

(P = 0,

Q = x,

(P = −y,

Q = 0,

8<:

P = −y

2,

Q =x

2.

Then, if we use Green’s Theorem in reverse we see that the area of the region R can also be computedby evaluating any of the following line integrals.

Area =

I

C

x dy = −I

C

y dx =1

2

I

C

x dy − y dx,

where C is the boundary curve of the region R.

Let us take a quick look at an example of this.


Use Green’s Theorem to find the area of a disk of radius a.

Solution We can use either of the integrals above, but the third one is probably the easiest. So,

area =1

2

I

C

x dy − y dx,

where C is the circle of radius a. In order to do this we need a parametrization of C. This is

x = a cos t, y = a sin t, 0 6 t 6 2π.

The area is then

area =1

2

I

C

x dy − y dx

=1

2

„Z 2π

0

a cos t (a cos t) dt−Z 2π

0

a sin t (−a sin t) dt

«

=1

2

Z 2π

0

`a2 cos2 t + a2 sin2 t

´dt

=1

2

Z 2π

0

a2 dt

= πa2.

Similarly, can you repeat the question and find the area of an ellipse of lengths a and b ? [ Ans: πab ]

2

196


• Curl and divergence

Here we are going to introduce a couple of new concepts, the curl and the divergence of a vector. Thepurpose of doing this is to introduce the vector form of Green’s Theorem.

Let us start with the curl. Given the vector field F = P i + Q j + Rk the curl is defined to be

curl F =

„∂R

∂y− ∂Q

∂z

«i +

„∂P

∂z− ∂R

∂x

«j +

„∂Q

∂x− ∂P

∂y

«k.

There is another (potentially) easier definition of the curl of a vector field but we will first need to definethe ∇ operator. This is defined to be

∇ =∂

∂xi +

∂

∂yj +

∂

∂zk.

We use this as if it is a function with another input function f such that ∇f = fx i + fy j + fz k. So,whatever function is listed after the ∇ is substituted into the partial derivatives. Note as well that whenwe look at it in this way we simply get the gradient vector.

Using the ∇ we can define the curl as the following cross product

curl F = ∇× F =

˛̨˛̨˛̨˛̨˛

i j k

∂

∂x

∂

∂y

∂

∂z

P Q R

˛̨˛̨˛̨˛̨˛.

We have a couple of nice facts that use the curl of a vector field.

Fact 5

1. If f(x, y, z) has continuous second-order partial derivatives then curl (∇f) = 0. This is easyenough to check by plugging into the definition of the derivative so we will leave it to you for anexercise.

2. If F is a conservative vector field then curl F = 0. This is a direct result of what it means to bea conservative vector field in the previous part.

3. If F is defined on all of R3 whose components have continuous first-order partial derivatives andcurl F = 0 then F is a conservative vector field. This is not so easy to verify and we will skipthe details of the proof here. However you can freely use this fact if it is applicable.

The above indeed provides a way of determining a given three-dimensional vector field is conservative ornot. (For two-dimensional case the condition is simply Qx = Py, page 185) The explicit conditions are

∂P

∂y=

∂Q

∂x,

∂P

∂z=

∂R

∂x,

∂Q

∂z=

∂R

∂y.

197


¥ Example 5.3.5 (Curl)

Determine if F = x2y i + xyz j− x2y2 k is a conservative vector field.

Solution So all that we need to do is compute the curl and see if we get the zero vector or not.

curl F =

˛̨˛̨˛̨˛̨˛

i j k

∂

∂x

∂

∂y

∂

∂z

x2y xyz −x2y2

˛̨˛̨˛̨˛̨˛

=

„∂(−x2y2)

∂y− ∂(xyz)

∂z

«i−„

∂(−x2y2)

∂x− ∂(x2y)

∂z

«j +

„∂(xyz)

∂x− ∂(x2y)

∂y

«k

=`−2x2y − xy

´i− `−2xy2´ j +

`yz − x2´k

6= 0.

The curl is not the zero vector and so the given vector field is not conservative. 2

Next we should talk about a physical interpretation of the curl. Suppose that F(x, y, z) is the velocity fieldof a flowing fluid. Then curl F represents the tendency of particles at the point (x, y, z) to rotate aboutthe axis that points in the direction of curl F. If curl F = 0 then the fluid is called irrotational.

Let us now discuss about the second new concept in the following. Given the three-dimensional vectorfield F = P i + Q j + Rk, the divergence is defined to be

div F =∂P

∂x+

∂Q

∂y+

∂R

∂z.

There is also a definition of the divergence in terms of the ∇ operator. The divergence can be defined interms of the following dot product.

div F = ∇ · F.

¥ Example 5.3.6 (Divergence)

Compute div F for F = x2y i + xyz j− x2y2 k.

Solution There really is not too much to do here other than compute the derivatives.

div F =∂(x2y)

∂x+

∂(xyz)

∂y+

∂(x2y2)

∂z= 2xy + xz.

2

We also have the following fact about the relationship between the curl and the divergence.

div“curl F

”= 0.

198


¥ Example 5.3.7 (Divergence of curl)

Verify the above fact for the vector field F(x, y, z) = yz2 i + xy j + yz k.

Solution Let us first compute the curl.

curl F =

˛̨˛̨˛̨˛̨˛

i j k

∂

∂x

∂

∂y

∂

∂z

yz2 xy yz

˛̨˛̨˛̨˛̨˛

= (z) i + (2yz) j +`y − z2´k.

Now compute the divergence of this.

div“curl F

”=

∂(z)

∂x+

∂(2yz)

∂y+

∂(y − z2)

∂z= 2z − 2z = 0.

2

We also have a physical interpretation of the divergence. If we again think of F as the velocity field of aflowing fluid then div F represents the net rate of change of the mass of the fluid flowing from the point(x, y, z) per unit volume. This can also be thought of as the tendency of a fluid to diverge from a point. Ifdiv F = 0 then the fluid is called incompressible.

The next topic that we want to briefly mention is the Laplace operator. Let us first take a look at

div“∇f”

= ∇ · ∇f = fxx + fyy + fzz.

The Laplace operator is then defined as

∇2 = ∇ · ∇.

The Laplace operator arises naturally in many fields including heat transfer and fluid flow.

The final topic in this section is to give two vector forms of Green’s Theorem. The first form uses thecurl of the vector field and is

I

C

F · dr =

ZZ

D

“curl F

”· k dA, or

I

C

F · dr =

ZZ

D

“∇× F

”· k dA,

where k is the standard unit vector in the positive z direction.

Remark that

curl F =

„∂R

∂y− ∂Q

∂z

«i +

„∂P

∂z− ∂R

∂x

«j +

„∂Q

∂x− ∂P

∂y

«k,

curl F · k =∂Q

∂x− ∂P

∂y.

So, the above first vector form of Green’s Theorem has absolutely no difference from the previous one inTheorem 5.3.1 (page 190), except we are using region D here instead of R.

199


The second form uses the divergence. In this case we also need the outward unit normal to the two-dimensional curve C. If the curve is parametrized by

r(t) = x(t) i + y(t) j,

then the outward unit normal is given by

n =y′(t)‖r′(t)‖ i− x′(t)

‖r′(t)‖ j.

Here is a sketch illustrating the outward unit normal for some curve C at various points.

Curve C

EnclosedRegion D

n

n

n

n

n

n

The vector form of Green’s Theorem that uses the divergence is given by

I

C

F · n ds =

ZZ

D

div F dA, or

I

C

F · n ds =

ZZ

D

∇ · F dA.

This form says that under suitable conditions the outward flux of a vector field across a simple closed curvein the plane equals the double integral of the divergence of the field over the region enclosed by the curve.If the vector field is F = P i + Q j, then the formula reduces to

I

C

F · n ds =

ZZ

D

„∂P

∂x+

∂Q

∂y

«dA.

The tiny directed circle on the line integral is a reminder that the integration around the closed curve C isto be in the counter-clockwise direction. To evaluate the line integral (or the flux integral, see words below)we do not need to know either n (although the formula for n is given above) or ds. All we need to dois evaluate the simple double integral on the right side.

At this point we should discuss what is the physical meaning of the line integral over C of F · n. IfF represents a fluid’s velocity field. Then F · n is the scalar component of the fluid’s velocity field inthe direction of the curve’s outward-pointing normal vector. To find the line integral

HC

F · n ds means

to find the rate at which a fluid is entering or leaving a region enclosed by a smooth closed curve C. Thevalue of this integral is the flux of F across C. For instance, the vector field F = −y i + x j is thevelocity field of a steady-state counter-clockwise rotation around the origin. Then the flux of F acrossany simple closed curve C is only zero! This follows immediately from Green’s Theorem because∇ · F = ∂

∂x(−y) + ∂

∂y(x) = 0.

200

5.4 Surface Integrals


In previous sections we considered three kinds of integrals – double integrals over two-dimensional regions,triple integrals over three-dimensional solids, and line integrals along curves in two- or three-dimensionalspace. In this section we discuss integrals over surfaces in three-dimensional space. Such integrals arecalled surface integrals. Surface integrals occur in physical problems involving fluid and heat flow,electricity, magnetism, mass, and center of gravity.

• Parametric surfaces

Before we get into surface integrals we first need to talk about how to parametrize a surface. When weparametrize a curve in 3-space we take values of t from some interval [a, b] and plug them into

r(t) = x(t) i + y(t) j + z(t)k,

and the resulting set of vectors will be the position vectors for the points on the curve.

With surfaces we will do something similar. We will take points (u, v) out of some two-dimensionalregion R and plug them into

r(u, v) = x(u, v) i + y(u, v) j + z(u, v)k,

and the resulting set of vectors will be the position vectors for the points on the surface S that we are tryingto parametrize. This is often called the parametric representation of the parametric surface S.The variables u, v are parameters, and R is the parameter domain.

We will sometimes need to write the parametric equations for a surface. There are really nothingmore than the three components of the parametric representation explicitly written down.

x = x(u, v), y = y(u, v), z = z(u, v).

¥ Example 5.4.1 (Parametric equations for surface)

Determine the surface given by the parametric representation

r(u, v) = u i + u cos v j + u sin v k.

Solution Let us first write down the parametric equations.

x = u, y = u cos v, z = u sin v.

Now if we square y and z and then add them together we get

y2 + z2 = u2 cos2 v + u2 sin2 v = u2 = x2.

So, we are able to eliminate the parameters and the equation for x, y, and z is given by

x2 = y2 + z2.

Interpret as a quadric surface we can see that this is a cone that opens along the x-axis. 2

In fact, we are much more likely to need to write down the parametric equations of a surface than identifythe surface from the parametric representation so let us take a look at some examples of this.

201


¥ Example 5.4.2 (Parametric equations for surface)

Give parametric representations for each of the following surfaces.

(a) The elliptic paraboloid x = 5y2 + 2z2 − 10.

(b) The elliptic paraboloid x = 5y2 + 2z2 − 10 that is in front of the yz-plane.

(c) The sphere x2 + y2 + z2 = 30.

(d) The cylinder y2 + z2 = 25.

Solution

(a) This one is probably the easiest one of the four to see how to do. Since the surface is in the formx = f(y, z) we can quickly write down a set of parametric equations as follows

x = 5y2 + 2z2 − 10, y = y, z = z.

The parametric representation is then

r(y, z) =`5y2 + 2z2 − 10

´i + y j + z k.

(b) This is really a restriction on the previous parametric representation. The parametric representationstays the same.

r(y, z) =`5y2 + 2z2 − 10

´i + y j + z k.

However, since we only want the surface that lies in front of the yz-plane we also need to require thatx > 0. This is equivalent to requiring

5y2 + 2z2 − 10 > 0 =⇒ 5y2 + 2z2 > 10.

(c) This one can be a little tricky until you see how to do it. In spherical coordinates we know that theequation of a sphere of radius a centered at the origin is given by

ρ = a

and so the equation of this sphere (in spherical coordinates) is ρ =√

30. Now, we also have thefollowing conversion formulas for converting Cartesian coordinates into spherical coordinates

x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ.

However, we know that ρ is for our sphere and so if we plug this into these conversion formulas wewill arrive at a parametric representation for the sphere. Therefore, the parametric representation is

r(φ, θ) =√

30 sin φ cos θ i +√

30 sin φ sin θ j +√

30 cos φ k.

All that we need to do now is come up with some restriction on the variables. First we know that wehave the following restriction.

0 6 φ 6 π.

This is enforced upon us by choosing to use spherical coordinates. Also, to make sure that we onlytrace out the sphere once we will also have the following restriction.

0 6 θ 6 2π.

202


(d) As with the last one this can be tricky until you see how to do it. In this case it makes some sense touse cylindrical coordinates since they can be easily used to write down the equation of a cylinder.

In cylindrical coordinates the equation of a cylinder of radius a is given by

r = a,

and so the equation of the cylinder in this problem is r = 5.

Next, we have the following conversion formulas

x = x, y = r cos θ, z = r sin θ.

Notice that they are slightly different from those that we are used to seeing. We need to change themup here since the cylinder is centered upon the x-axis.

The parametric representation for the cylinder is then

r(x, θ) = x i + 5 cos θ j + 5 sin θ k.

We will also need the restriction 0 6 θ 6 2π to make sure that we do not retrace any portion of thecylinder. Since we have not put any restrictions on the “height” of the cylinder there won’t be anyrestriction on x.

2

In part (a) of this example we used the fact that the function is in the form x = f(y, z) to quickly writedown a parametric representation. This can always be done for functions that are in this basic form.

z = f(x, y) =⇒ r(x, y) = x i + y j + f(x, y)k,

x = f(y, z) =⇒ r(y, z) = f(y, z) i + y j + z k,

y = f(x, z) =⇒ r(x, z) = x i + f(x, z) j + z k.

Now we have practice writing down some parametric representations for some surfaces let us take a quicklook at a couple of applications.

The first application is to find the tangent plane to the parametric surface S given by

r(u, v) = x(u, v) i + y(u, v) j + z(u, v)k.

We need S to be smooth for the construction we are about to carry out. The definition of smoothnessinvolves the partial derivatives of r with respect to u and v:

ru(u, v) =∂x

∂u(u, v) i +

∂y

∂u(u, v) j +

∂z

∂u(u, v)k,

rv(u, v) =∂x

∂v(u, v) i +

∂y

∂v(u, v) j +

∂z

∂v(u, v)k.

A parametrized surface r(u, v) = x(u, v) i+ y(u, v) j+ z(u, v)k is smooth if ru and rv are continuousand ru × rv 6= 0 on the interior of the parameter domain. The condition that ru × rv is never the zerovector means that the two vectors are nonzero and never lie along the same line, so they always determinea plane tangent to the surface. In other words, the vector ru × rv will be orthogonal to the surface S.This means that it can be used for the normal vector that we need in order to write down the equation of atangent plane.

Let us take a look at an example.

203


¥ Example 5.4.3 (Tangent plane to surface)

Find the equation of the tangent plane to the surface given by

r(u, v) = u i + 2v2 j + (u2 + v)k

at the point (2, 2, 3).

Solution Let us first compute ru × rv. Here are the two individual vectors.

ru = i + 2uk, rv = 4v j + k.

Now the cross product (which will give us the normal vector n) is

n = ru × rv =

˛̨˛̨˛̨˛̨

i j k

1 0 2u

0 4v 1

˛̨˛̨˛̨˛̨

= −8uv i− j + 4v k.

In order to use this normal vector we will need to determine the value of u and v that will give us the pointin the question. We can easily do this by setting the individual components of the parametric representationequal to the coordinates of the point in question. Doing this gives

u = 2 =⇒ u = 2,

2v2 = 2 =⇒ v = ±1,

u2 + v = 3.

As shown above we have the value of u but there are two values of v. To determine the correct value ofv let us plug u into the third equation and solve for v. This should tell us what the correct value is.

4 + v = 3 =⇒ v = −1.

We now know that we will be at the point in question when u = 2 and v = −1. At this point the normalvector is

n = 16i− j− 4k.

The tangent plane is then

16(x− 2)− (y − 2)− 4(z − 3) = 0,

16x− y − 4z = 18.

2

The second application that we want to take a quick look at is the surface area of the parametric surfaceS given by


Provided S is traced out exactly once as (u, v) ranges over the points in R such that a 6 u 6 b,c 6 v 6 d, the surface area of S is given by

Surface Area =

ZZ

R

‖ru × rv‖ dA =

Z d

c

Z b

a

‖ru × rv‖ du dv.

204


We can abbreviate the integral in the above by writing dσ for ‖ru×rv‖ du dv. The surface area differentialdσ is analogous to the arc length differential ds in Section 5.2 (page 169).

Surface Area =

ZZ

S

1 dσ, dσ = ‖ru × rv‖ du dv.

Let us give a quick explanation of how we get this formula for finding surface area. Remark that weindeed approximated the surface area element ∆σ by the parallelogram on the tangent plane whose sidesare determined by the vectors ∆u ru and ∆v rv. The area of the parallelogram is

‖∆u ru ×∆v rv‖ = ‖ru × rv‖ ∆u ∆v.

Summing these elemental areas together to obtain an approximation of the surface area of S:

X‖ru × rv‖ ∆u ∆v ≈

Z d

c

Z b

a

‖ru × rv‖ du dv,

where ∆u, ∆v are taken to approach zero. The double integral over region R defines the area of thesurface S.

Let us take a look at a couple of examples.

¥ Example 5.4.4 (Surface area)

Find the surface area of the cone

z =p

x2 + y2, 0 6 z 6 1.

Solution With reference to Example 5.4.1 (page 201) the given cone (opens along the positive z-axis)has a parametrization (taking u = r and v = θ)

r(r, θ) = (r cos θ) i + (r sin θ) j + r k, 0 6 r 6 1, 0 6 θ 6 2π.

To apply the equation for finding surface area, we first find rr × rθ :

rr × rθ =

˛̨˛̨˛̨˛̨

i j k

cos θ sin θ 1

−r sin θ r cos θ 0

˛̨˛̨˛̨˛̨

= −r cos θ i− r sin θ j + (r cos2 θ + r sin2 θ)k

= −r cos θ i− r sin θ j + r k.

Thus, ‖rr × rθ‖ =p

r2 cos2 θ + r2 sin2 θ + r2 =√

2r2 =√

2r. The surface area of the cone is then

surface area =

Z 2π

0

Z 1

0

‖rr × rθ‖ dr dθ

=

Z 2π

0

Z 1

0

√2r dr dθ

=√

2 · (2π) (1

2) = π

√2 units squared.

2

205


¥ Example 5.4.5 (Surface area)

Find the surface area of a sphere of radius a.

Solution First we need a parametrization of the sphere. Spherical coordinates provide what we need. Atypical point (x, y, z) on the sphere has

x = a sin φ cos θ, y = a sin φ sin θ, and z = a cos φ,

where

0 6 φ 6 π, 0 6 θ 6 2π.

Taking u = φ and v = θ gives the parametrization

r(φ, θ) = (a sin φ cos θ) i + (a sin φ sin θ) j + (a cos φ)k, 0 6 φ 6 π, 0 6 θ 6 2π.

To apply the equation for finding surface area, we find rφ × rθ :

rφ × rθ =

˛̨˛̨˛̨˛̨

i j k

a cos φ cos θ a cos φ sin θ −a sin φ

−a sin φ sin θ a sin φ cos θ 0

˛̨˛̨˛̨˛̨

= (a2 sin2 φ cos θ) i + (a2 sin2 φ sin θ) j + (a2 sin φ cos φ)k.

Thus,

‖rφ × rθ‖ =p

a4 sin4 φ cos2 θ + a4 sin4 φ sin2 θ + a4 sin2 φ cos2 φ

=

qa4 sin4 φ + a4 sin2 φ cos2 φ

=q

a4 sin2 φ`sin2 φ + cos2 φ

´

= a2q

sin2 φ = a2 sin φ,

since sin φ > 0 for 0 6 φ 6 π. Therefore, the area of the sphere is

surface area =

Z 2π

0

Z π

0

‖rφ × rθ‖ dφ dθ

=

Z 2π

0

Z π

0

a2 sin φ dφ dθ

= (2π)h− a2 cos φ

iπ

0

= (2π)(2a2) = 4πa2 units squared.

This agrees with the well-known formula for the surface area of a sphere. 2

206


• Surface integrals

It is now time to think about integrating functions over some surface, S, in three-dimensional space. Letus start with a sketch of the surface S since the notation can get a little confusing once we get into it.Here is a sketch of some surface S.

x

y

z

Surface S

Region R��

��

The surface S will lie above (in this case) some region R that lies in the xy-plane. We use a rectanglehere, but it doesn’t have to be of course. Also note that we can just as easily look at a surface S that is infront of some region R in the yz-plane or the xz-plane. Do not get so locked into the xy-plane that youcannot do problems that have regions in the other two planes. Now, how we evaluate the surface integralwill depend upon how the surface is given to us. There are essentially two separate methods here, althoughas we will see they are really the same.

First, let us look at the surface integral in which the surface S is given by z = g(x, y). In this casethe surface integral is

ZZ

S

f(x, y, z) dσ =

ZZ

R

f(x, y, g(x, y))

s“∂g

∂x

”2

+“∂g

∂y

”2

+ 1 dA.

Now, we need to be careful here as both of these look like standard double integrals. In fact the integral onthe right is a standard double integral. The integral on the left however is a surface integral. The way totell the difference is by looking at the differentials. The surface integral will have a surface area differentialdσ while the standard double integral will have a dA.

In order to evaluate a surface integral we will substitute the equation of the surface in for z in theintegrand and then add on the often messy square root. After that the integral is a standard double integraland by this point we should be able to deal with that. Note as well that there are similar formulas forsurfaces given by y = g(x, z) (with R in the xz-plane) and x = g(y, z) (with R in the yz-plane). Wewill see one of these formulas in the examples.

To motivate the definition of a surface integral we will consider the problem of finding the mass of acurved lamina whose density function (mass per unit area) is known. If S models a lamina, we may letf(x, y, z) denote the density function, then the mass of the lamina is given by M =

RRS

f(x, y, z) dσ. Note

also that if f is identically 1, then the formula will give the surface area =RRS

1 dσ.

207


The second method for evaluating a surface integral is for those surfaces that are given by the vectorequivalence of parametrization


In these cases the surface integral is

ZZ

S

f(x, y, z) dσ =

ZZ

R

f(r(u, v)) ‖ru × rv‖ dA,

where R is the range of the parameters that trace out the surface S.

Before we work on examples let us notice that since we can parametrize a surface given by z = g(x, y) as

r(x, y) = x i + y j + g(x, y)k,

we can always use this form for these kind of surfaces as well. In fact it can be shown that

‖rx × ry‖ =

s“∂g

∂x

”2

+“∂g

∂y

”2

+ 1

for these kinds of surfaces. You might want to verify this for the practice of computing these cross product.Let us work some examples.

¥ Example 5.4.6 (Surface integrals)

Evaluate

ZZ

S

6xy dσ where S is the portion of the plane x + y + z = 1 in the first octant that lies in

front of the yz-plane.

Solution Since we are looking for the portion of the plane in the first octant that lies in front of theyz-plane we are going to write the equation of the surface in the form x = g(y, z). This is easy enough.

x = 1− y − z.

Next we need to determine just what R is. Here is a sketch of the surface S and the region R.

z

y

x

0

1

1

1

Region Ron yz-plane

Surface Sx + y + z = 1

��

�� z

y

Region R

0

1

1

z = 1− y

��

��

208


We obtain the equation of the hypotenuse of the triangular region R by setting x equal to zero in theequation of the plane and solving for z. Here are the ranges for y and z.

0 6 y 6 1, 0 6 z 6 1− y.

Now, because the surface is not in the form z = g(x, y) we cannot directly use the formula above. However,as noted before we can modify this formula to get one that will work for us.

ZZ

S

f(x, y, z) dσ =

ZZ

R

f(g(y, z), y, z)

s1 +

“∂g

∂y

”2

+“∂g

∂z

”2

dA.

The changes analogously made to the formula should be somewhat obvious. So let us do the integral.ZZ

S

6xy dσ =

ZZ

R

6(1− y − z)yp

1 + (−1)2 + (−1)2 dA.

Notice that we plug in the equation of the plane for the x in the integrand. At this moment we have afairly simple double integral to do. Here is the work.

ZZ

S

6xy dσ =√

3

ZZ

R

6(y − y2 − yz) dA

= 6√

3

Z 1

0

Z 1−y

0

(y − y2 − yz) dz dy

= 6√

3

Z 1

0

»yz − y2z − 1

2yz2

–1−y

0

dy

= 6√

3

Z 1

0

„1

2y − y2 +

1

2y3

«dy

= 6√

3

»1

4y2 − 1

3y3 +

1

8y4

–1

0

=

√3

4.

2


Evaluate

ZZ

S

z dσ where S is the upper half of a sphere of radius 2.

Solution The parametrization of a general sphere of radius a is given in Example 5.4.5 (page 206). Theparametrization of the upper half of the sphere of radius 2 is

r(φ, θ) = (2 sin φ cos θ) i + (2 sin φ sin θ) j + (2 cos φ)k, 0 6 φ 6 π

2, 0 6 θ 6 2π.

By the result of Example 5.4.5 we know that

‖rφ × rθ‖ = 4 sin φ.

Recall that the surface integral can be evaluated asZZ

S

f(x, y, z) dσ =

ZZ

R

f(r(φ, θ)) ‖rφ × rθ‖ dA.

209


Do not forget that we need to plug in for x, y, and/or z in the integrand as well, although in this casewe just have to plug in z. Here is the evaluation for the double integral.

ZZ

S

z dσ =

Z 2π

0

Z π/2

0

(2 cos φ) (4 sin φ) dφ dθ

= (2π)

Z π/2

0

4 sin 2φ dφ

= (2π)h− 2 cos 2φ

iπ/2

0= 8π.

2


Evaluate

ZZ

S

y dσ where S is the portion of the cylinder x2 + y2 = 3 that lies between z = 0 and

z = 6.

Solution In terms of cylindrical coordinates we can find a parametrization of the cylinder. Here it is.

r(z, θ) =√

3 cos θ i +√

3 sin θ j + z k.

The ranges for the parameters are

0 6 z 6 6, 0 6 θ 6 2π.

Now we need the cross product rz × rθ. Here are the two vectors.

rz = k, rθ = −√

3 sin θ i +√

3 cos θ j.

Hence the cross product is

rz × rθ =

˛̨˛̨˛̨˛̨

i j k

0 0 1

−√3 sin θ√

3 cos θ 0

˛̨˛̨˛̨˛̨

= −√

3 cos θ i−√

3 sin θ j.

The magnitude of this vector is

‖rz × rθ‖ =p

3 cos2 θ + 3 sin2 θ =√

3.

The surface integral is then

ZZ

S

y dσ =

ZZ

R

(√

3 sin θ) (√

3) dA

= 3

Z 2π

0

Z 6

0

sin θ dz dθ

= 3

Z 2π

0

6 sin θ dθ

=h− 18 cos θ

i2π

0= 0.

2

210


• Surface integrals of vector fields

Just as we did with line integrals we now need to move on to surface integrals of vector fields. Recall that inline integrals the orientation of the curve we are integrating along can change the answer. The same thingwill hold true with surface integrals. So, before we really get into doing surface integrals of vector fields wefirst need to introduce the idea of an orientable surface.

Most surfaces that we encounter in applications have two sides – a sphere has an inside and an outside,and an infinite horizontal plane has a top and a bottom side, for example. This may seem strange, thereexist mathematical surfaces with only one side, recall that the Mobius strip is a surface that only has oneside! Anyway, let us start with a surface that has two sides. A two-sided surface is said to be orientable.In the rest of this text we will only be concerned with orientable surfaces.

In applications, it is important to have some way of distinguishing between the two sides of an orientablesurface. For this purpose let us suppose that S is an orientable surface that has a unit normal vector nat each point (except possibly along the boundary). More precisely, every point will have two unit normalvectors, n1 and n2 = −n1. The vectors n1 and n2 point to opposite sides of the surface and hence serveto distinguish between the two sides. These unit normal vectors are then said to form an orientation ofthe surface. It can also be proved that a smooth orientable surface has only two possible orientations. Onceyou fix an orientation the surface is called an oriented surface.

There is one convention that we will make in regards to oriented surfaces. First we need to define aclosed surface. A surface S is closed if it is the boundary of some solid G. A good example of a closedsurface is the surface of a sphere. We say that the closed surface S has a positive orientation if wechoose the set of unit normal vectors that point outward from the solid G while the negative orientationwill be the set of unit normal vectors that point in towards the solid G. Note that the convention is onlyused for closed surfaces.

In order to work with surface integrals of vector fields we will need to be able to write down a formulafor the unit normal vector corresponding to the orientation that we have chosen to work with. We have twoways of doing this depending on how the surface has been given to us.

First, let us suppose that the function is given by z = g(x, y). In this case we first define a new function

f(x, y, z) = z − g(x, y).

In terms of our new function the surface is then given by the equation f(x, y, z) = 0. Now, recall that∇f will be the orthogonal (or normal) to the surface given by f(x, y, z) = 0. This means that we have anormal vector to the surface. The only potential problem is that it may not be a unit normal vector. Thatis not a problem since we also know that we can turn any nonzero vector into a unit vector by dividing thevector by its length. Thus we have

n =∇f

‖∇f‖ .

In this case it will be more convenient to actually compute the gradient vector and plug this into the formulafor the normal vector. Doing this gives

n =∇f

‖∇f‖ =−gx i− gy j + kp(gx)2 + (gy)2 + 1

.

Now, from a notational standpoint this may not be so convenient, but it does allow us to make a coupleadditional comments.

211


First, notice that the component of the normal vector in the z-direction (identified by the k in thenormal vector) is always positive and so this normal vector will generally point upwards. It may not pointdirectly up, but it will have an upwards component to it. This will be important when we are working witha closed surface and we want the positive orientation. If we know that we can then look at the normalvector and determine if the “positive” orientation should point upwards or downwards. Remember that the“positive” orientation must point out of the solid and this may mean downwards in places. Of course if itturns out that we need the downward orientation we can always take the negative of this unit vector andwe will get the one that we need. Again, remember that we always have that option when choosing the unitnormal vector.

Before we move onto the second method of giving the surface we should point out that we only do thisfor surfaces in the form z = g(x, y). We can just as easily do the above work for surfaces in the formy = g(x, z) (so f(x, y, z) = y−g(x, z) ) or for surfaces in the form x = g(y, z) (so f(x, y, z) = x−g(y, z) ).

We need to discuss how to find the unit normal vector if the surface is given parametrically as


In this case recall (page 203) that the vector ru × rv will be normal to the tangent plane at a particularpoint. But if the vector is normal to the tangent plane at a point then it will also be normal to the surfaceat that point. So, this is a normal vector. In order to guarantee that it is a unit normal vector we will alsoneed to divide it by its magnitude.

So, in the case of parametric surfaces one of the unit normal vectors will be

n =ru × rv

‖ru × rv‖ .

As with the first case we will need to look at this once it is computed and determine if it points in the correctdirection or not. If it does not then we can always take the negative of this vector and that will point inthe correct direction. Finally, remember that we can always parametrize any surface given by z = g(x, y)(or y = g(x, z) or x = g(y, z) ) easily enough and so if we want to we can always use the parametrizationformula to find the unit normal vector.

Now that we have looked at orientable surfaces and their associated unit normal vectors we can actuallygive a formula for evaluating surface integrals of vector fields. Given a vector field F with unit normalvector n then the surface integral of F over the surface S is given by

ZZ

S

F · n dσ,

where the right hand integral is a standard surface integral. This is sometimes called the flux of F acrossthe surface S.

In Physics this is a measure of representing the exact net volume of fluid that crosses the surface Sin the direction of its orientation n per unit of time. A positive flux means that in one unit of time agreater volume of fluid passes through S in the positive direction than in the negative direction, a negativeflux means that a greater volume passes through the surface in the negative direction than in the positivedirection, and a zero flux means that the same volume passes through the surface in each direction. Integralsof this form are called flux integrals.

212


Before we work any examples let us notice that we can substitute in for the unit normal vector to geta somewhat easier formula to use. We will need to be careful with each of the following formulas howeveras each will assume a certain orientation and we may have to change the normal vector to match the givenorientation.

Let us first start by assuming that the surface is given by z = g(x, y). In this case let us also assumethat the vector field is given by F = P i + Q j + Rk and that the orientation that we are after is the“upward” orientation. Under all of these assumptions the surface integral of F over S is

ZZ

S

F · n dσ =

ZZ

D

(P i + Q j + Rk) ·

−gx i− gy j + kp(gx)2 + (gy)2 + 1

! p(gx)2 + (gy)2 + 1 dA

=

ZZ

D

(P i + Q j + Rk) · (−gx i− gy j + k) dA

=

ZZ

D

(−Pgx −Qgy + R) dA.

Now, remember that this assumes the “upward” orientation. If we need the “downward” orientation thenwe need to change the signs on the normal vector. This would in turn change the signs on the integrandas well. So, we really need to be careful here when using this formula. In general it is best to rederive thisformula as you need it.

When we have been given a surface that is not in parametric form there are in fact 6 possible integralshere. Two for each form of the surface z = g(x, y), y = g(x, z) and x = g(y, z). Given each form of thesurface there be two possible unit normal vectors and we will need to choose the correct one to match thegiven orientation of the surface. However, the derivation of each formula is similar to that given here andso should not be too difficult to do as you need to.

Notice as well that because we are using the unit normal vector the messy square root will always dropout. This means that when we do need to derive the formula we won’t really need to put this in. All wewill need to work with is the numerator of the unit vector. We will see at least one more of these derived inthe examples below. It should also be noted that the square root is nothing more than

p(gx)2 + (gy)2 + 1 = ‖∇f‖

so in the following work we will probably just use the notation in place of the square root when we can makethings a little simpler.

Let us now take a quick look at the formula for the surface integral when the surface is given paramet-rically by r(u, v). In this case the surface integral is

ZZ

S

F · n dσ =

ZZ

D

F ·„

ru × rv

‖ru × rv‖«‖ru × rv‖ dA

=

ZZ

D

F · (ru × rv) dA.

Again note that we may have to change the sign on ru × rv to match the orientation of the surface and sothere is once again really two formulas here. Also note that again the magnitude cancels in this case and sowe won’t need to worry that in these problems either. Let us now work a couple of examples.

213



EvaluateZZ

S

F · n dσ,

where F = y j− z k and S is the surface given by the paraboloid y = x2 + z2, 0 6 y 6 1 and the diskx2 + z2 6 1 at y = 1. Assume that S has positive orientation.

Solution First let us notice that the disk is really nothing more than the cap on the paraboloid. Thismeans that we have a closed surface. This is important because we have been told that the surface has apositive orientation and by convection this means that all the unit normal vectors will need to point outwardsfrom the solid enclosed by S.

Let us first get a sketch of S so we can get a feel for what is going on and in which direction we willneed the unit normal vectors to point.

x

y

z

S2

S1

��

��

As noted in the sketch we will denote the paraboloid S1 and the disk by S2. Also note that in order forunit normal vectors on the paraboloid to point away from the solid they will all need to point generally inthe negative y direction in order to point away from the solid.

Since S is composed of two surfaces we will need to do the surface integral on each and then add theresults to get the overall surface integral. Let us start with the paraboloid. In this case we have the surfacein the form y = g(x, z) so we will need to derive the correct formula since the one given initially is not forthis kind of function. This is easy enough to do however. First define

f(x, y, z) = y − g(x, z) = y − x2 − z2.

We will next need the gradient vector of this function.

∇f = 〈−2x, 1,−2z〉.

214


Now, the y component of the gradient is positive and so this vector will generally point in the positive ydirection. However, as noted before we need the normal vector point in the negative y direction to makesure that it will be pointing away from the enclosed solid. This means that we will need to use

n =−∇f

‖ − ∇f‖ =〈2x,−1, 2z〉‖∇f‖ .

Let us note a couple of things here before we proceed. We do not really need to derive this by the magnitudeof the gradient since this will just cancel out once we actually do the integral. So, because of this we will notcompute the magnitude ‖∇f‖. Also, the dropping of the minus sign inside the magnitude is not a typo.When we compute the magnitude we are going to square each of the components and so the minus sign willdrop out.

Here is the surface integral on the paraboloid.

ZZ

S1

F · n dσ =

ZZ

D

(y j− z k) ·„ 〈2x,−1, 2z〉

‖∇f‖«‖∇f‖ dA

=

ZZ

D

`−y − 2z2´ dA

=

ZZ

D

ˆ−(x2 + z2)− 2z2˜ dA

= −ZZ

D

`x2 + 3z2´ dA.

Do not forget that we need to plug in the equation of the surface for y before we actually compute theintegral. In this case D is the disk of radius 1 in the xz-plane and so it makes sense to use polar coordinatesto compute the integral. Here are the polar coordinates for this region.

x = r cos θ, z = r sin θ, 0 6 θ 6 2π, 0 6 r 6 1.

Remark that the second formula is z = r sin θ instead of y = r sin θ since the disk is in the xz-plane.The following is the evaluation of this integral.

ZZ

S1

F · n dσ = −ZZ

D

`x2 + 3z2´ dA

= −Z 2π

0

Z 1

0

`r2 cos2 θ + 3r2 sin2 θ

´r dr dθ

= −Z 2π

0

Z 1

0

`cos2 θ + 3 sin2 θ

´r3 dr dθ

= −Z 2π

0

»1

2(1 + cos 2θ) +

3

2(1− cos 2θ)

– »1

4r4

–1

0

dθ

= −1

8

Z 2π

0

(4− 2 cos 2θ) dθ

= −1

8

h4θ − sin 2θ

i2π

0= −π.

215


We can now do the surface integral on the disk (cap on the paraboloid). This one is actually fairly easy todo and in fact we can use the definition of the surface integral directly. First let us notice that the disk isreally just the portion of the plane y = 1 that is in front of the disk of radius 1 in the xz-plane.

Now we want the unit normal vector to point away from the enclosed solid and since it must also beorthogonal to the plane y = 1 then it must point in a direction that is parallel to the y-axis, but we alreadyhave a known unit vector that does this, namely

n = j,

the standard unit basis vector. It also points in the correct direction for us to use. Because we have thevector field and the normal vector we can plug directly into the definition of the surface integral to get

ZZ

S2

F · n dσ =

ZZ

S2

(y j− z k) · j dσ =

ZZ

S2

y dσ.

Now we need to plug in for y (since S2 is a portion of the plane y = 1 we do know what it is) and wewill also need the square root this time when we convert the surface integral over to a double integral. Inthis case since we are using the definition directly we will not get the canceling of the square root that wesaw with the first portion. To get the square root we will need to acknowledge that

y = 1 = g(x, z)

and so the square root is

p(gx)2 + 1 + (gz)2.

The surface integral is then

ZZ

S2

F · n dσ =

ZZ

S2

y dσ =

ZZ

D

(1)√

0 + 1 + 0 dA =

ZZ

D

1 dA.

At this moment we can acknowledge that D is a disk of radius 1 and this double integral is nothing morethan the double integral that will give the area of the region D so there is no reason to compute theintegral. Here is the value of the surface integral.

ZZ

S2

F · n dσ = π.

Finally, to finish this we just need to add the two parts up. Here is the surface integral that we actuallyneed to compute.

ZZ

S

F · n dσ =

ZZ

S1

F · n dσ +

ZZ

S2

F · n dσ = −π + π = 0.

2

216



EvaluateZZ

S

F · n dσ,

where F = x i+y j+z4 k and S is the upper half of the sphere x2 +y2 +z2 = 9 and the disk x2 +y2 6 9in the plane z = 0. Assume that S has the positive orientation.

Solution As with the previous problem we have a closed surface and since we are told that the surfacehas a positive orientation all the unit normal vectors must point away from the enclosed surface. To helpus visualize this here is a sketch of the surface.

x

y

z

3

S1

S2

��

��

We will call S1 the hemisphere and S2 will be the bottom of the hemisphere (shown on the sketch). Now,in order for the unit normal vectors on the sphere to point away from enclosed surface they will all need tohave a positive z component. Remember that the vector must be normal to the surface and if there is apositive z component and the vector is normal it will have to be pointing away from the enclosed surface.On the other hand, the unit normal on the bottom of the disk must point in negative z direction in orderto point away from the enclosed surface.

Let us do the surface integral on S1 first. In this case since the surface is a sphere we will need to usethe parametric representation of the surface. This is

r(φ, θ) = 3 sin φ cos θ i + 3 sin φ sin θ j + 3 cos φ k.

Since we are working on the semisphere here are the limits on the parameters that we will need to use.

0 6 θ 6 2π, 0 6 φ 6 π

2.

Next, we need to determine rφ × rθ. Here are the two individual vectors.

rφ(φ, θ) = 3 cos φ cos θ i + 3 cos φ sin θ j− 3 sin φ k,

rθ(φ, θ) = −3 sin φ sin θ i + 3 sin φ cos θ j.

I should point out that it is completely possible for someone to write r(θ, φ) instead of r(φ, θ) here. Butthen naturally you will compute rθ×rφ instead of rφ×rθ. That will make a difference in the constructionof the normal vector n. We will investigate this again in the following. We first continue with our case.

217


The cross product is

rφ × rθ =

˛̨˛̨˛̨˛̨

i j k

3 cos φ cos θ 3 cos φ sin θ −3 sin φ

−3 sin φ sin θ 3 sin φ cos θ 0

˛̨˛̨˛̨˛̨

= 9 sin2 φ cos θ i + 9 sin2 φ sin θ j + 9 sin φ cos φ cos2 θ k + 9 sin φ cos φ sin2 θ k

= 9 sin2 φ cos θ i + 9 sin2 φ sin θ j + 9 sin φ cos φ k.

Note that we do not need the magnitude of the cross product since that will cancel out once we start doingthe integral. Also note that for the range of φ mentioned before both sine and cosine are positive and sothis vector will have a positive z component and as we noted in order for this to point away from the surfaceS1 we will need the z component to be positive. Therefore we will need to use the following vector for theunit normal vector.

n =rφ × rθ

‖rφ × rθ‖ =9 sin2 φ cos θ i + 9 sin2 φ sin θ j + 9 sin φ cos φ k

‖rφ × rθ‖ .

Recall that if we started with finding rθ × rφ in the beginning, we should take n = − rθ × rφ

‖rθ × rφ‖ instead.

Now we will drop the magnitude once we get to actually doing the integral since it will just cancel inthe integral. Next we will need

F“r(φ, θ)

”= 3 sin φ cos θ i + 3 sin φ sin θ j + 81 cos4 φ k.

We may get the dot product as well.

F“r(φ, θ)

” · (rφ × rθ) = 27 sin3 φ cos2 φ + 27 sin3 φ sin2 φ + 729 sin φ cos5 φ

= 27 sin3 φ + 729 sin φ cos5 φ.

Now we can do the integral.

ZZ

S1

F · n dσ =

ZZ

S1

F · (rφ × rθ) dA

=

Z 2π

0

Z π/2

0

`27 sin3 φ + 729 sin φ cos5 φ

´dφ dθ

=

Z 2π

0

Z π/2

0

`27 sin φ(1− cos2 φ) + 729 sin φ cos5 φ

´dφ dθ

= −Z 2π

0

»27

„cos φ− 1

3cos3 φ

«+

729

6cos6 φ

–π/2

0

dθ

=

Z 2π

0

279

2dθ = 279π.

218


Now, we need to do the integral over the bottom of the hemisphere. In this case we are looking at the diskx2 + y2 6 9 that lies in the plane z = 0 and so the equation of this surface is actually z = 0. The diskis really the region that tells us how much of the surface we are going to use. This also means that we canuse the definition of the surface integral here with

n = −k.

We need the negative sign since it must point away from the enclosed surface. The surface integral in thiscase is then

ZZ

S2

F · n dσ =

ZZ

S2

`x i + y j + z4 k

´ · (−k) dσ =

ZZ

S2

(−z4) dσ.

However, remember that we are in the plane given by z = 0 and so the surface integral becomes

ZZ

S2

F · n dσ =

ZZ

S2

(−z4) dσ =

ZZ

S2

0 dσ = 0.

The last step is to add the two pieces up. Here is the surface integral that we are looking for.

ZZ

S

F · n dσ =

ZZ

S1

F · n dσ +

ZZ

S2

F · n dσ = 279π + 0 = 279π.

2

Recall that if F is the velocity field of a fluid then the surface integral

ZZ

S

F · n dσ represents the net

amount of fluid flowing through the surface S, or so-called the flux of F across S.

¥ Example 5.4.11 (Flux integrals)

Let S be the portion of the surface z = 1− x2 − y2 that lies above the xy-plane, and suppose that S isoriented up. Find the flux of the vector field F (x, y, z) = x i + y j + z k across S.

Solution The flux Φ is given by

Φ =

ZZ

S

F · n dσ =

ZZ

R

F ·„− ∂z

∂xi− ∂z

∂yj + k

«dA

=

ZZ

R

(x i + y j + z k) · (2x i + 2y j + k) dA

=

ZZ

R

`x2 + y2 + 1

´dA

`since z = 1− x2 − y2´

=

Z 2π

0

Z 1

0

`r2 + 1

´r dr dθ

=

Z 2π

0

„3

4

«dθ =

3π

2.

2

219


5.5 Divergence Theorem

In this section we will be concerned with flux across surfaces, such as spheres, that “enclose” a region ofspace. We will show that the flux across such surfaces can be expressed in terms of the divergence of thevector field, and we will use this result to give a physical interpretation of the concept of divergence. In factwe are going to relate surface integrals to triple integrals.

In the last section we studied flux across general surfaces. Here we will be concerned exclusively withsurfaces that are boundaries of finite solids – the surface of a solid sphere, the surface of a solid box, orthe surface of a solid cylinder, for example. Such surfaces are said to be closed. A closed surface mayor may not be smooth, but most of the surfaces that arise in applications are piecewise smooth; thatis, they consist of finitely many smooth surfaces joined together at the edges (a box, for example). We willlimit our discussion to piecewise smooth surfaces that can be assigned an inward orientation (towardthe interior of the solid) and an outward orientation (away from the interior). It is very difficult tomake this concept mathematically precise, but the basic idea is that each piece of the surface is orientable,and oriented pieces fit together in such a way that the entire surface can be assigned an orientation.

In Section 5.3 (page 197) we defined the divergence of a vector field


as

div F =∂P

∂x+

∂Q

∂y+

∂R

∂z

but we did not give a clear physical explanation of its meaning at that time. The following result, knownas the Divergence Theorem or Gauss’s Theorem, will provide us with a physical interpretation ofdivergence in the context of fluid flow.

Theorem 5.5.1 (Divergence Theorem) Let G be a solid whose boundary surface S is orientedoutward (with positive orientation). Let F be a vector field whose components have continuous first-orderpartial derivatives and n is the outward unit normal on S. Then

ZZ

S

F · n dσ =

ZZZ

G

div F dV.

In words, the Divergence Theorem states:

The flux of a vector field across a closed surface with outward orientation is equal to the triple integralof the divergence over the region enclosed by the surface.

This is sometimes called the outward flux across the surface. Sometimes it is easier to find the fluxacross a closed surface by using the Divergence Theorem than by evaluating the flux integral directly. Thisis illustrated in the following example.

¥ Example 5.5.1 (Divergence theorem)

Use the Divergence Theorem to find the outward flux of the vector field F(x, y, z) = z k across the spherex2 + y2 + z2 = a2.

220

5.5 Divergence Theorem

Solution Let S denote the outward-oriented spherical surface and G the solid that it encloses. Thedivergence of the vector field is

div F =∂

∂z(z) = 1.

So, the flux across S is given by

Φ =

ZZ

S

F · dσ =

ZZZ

G

1 dV = volume of G =4πa3

3.

2

The Divergence Theorem is usually the method of choice for finding the flux across closed piecewisesmooth surfaces with multiple sections, since it eliminates the need for a separate integral evaluation overeach section. This is illustrated in the next couple of examples.


Use the Divergence Theorem to find the outward flux of the vector field F(x, y, z) = 2x i + 3y j + z2 kacross the unit cube in the following figure.

x

y

z

Solid G

n

nn

n

n

Solution Let S denote the outward-oriented surface of the cube and G the solid that it encloses. Thedivergence of the vector field is

div F =∂

∂x(2x) +

∂

∂y(3y) +

∂

∂z(z2) = 5 + 2z.

So, the flux across S is given by

Φ =

ZZ

S

F · dσ =

ZZZ

G

(5 + 2z) dV =

Z 1

0

Z 1

0

Z 1

0

(5 + 2z) dz dy dx

=

Z 1

0

Z 1

0

h5z + z2

i10dy dx =

Z 1

0

Z 1

0

6 dy dx = 6.

Instead of directly finding the flux by summing 6 surface integrals, Divergence Theorem helps to convertit to a single triple integral. 2

221



Use the Divergence Theorem to evaluateZZ

S

F · n dσ

where F = xy i− 12y2 j + z k and the surface consists of the three surfaces, z = 4− 3x2 − 3y2, 1 6 z 6 4

on the top, x2 + y2 = 1, 0 6 z 6 1 on the sides and z = 0 on the bottom.

Solution Let us start with the sketch of the surface.

x

y

z

4

1

1

n

n

z = 4− 3x2 − 3y2

x2 + y2 = 1

The solid G for the triple integral is then the solid enclosed by these surfaces. Note that cylindricalcoordinates will be a perfect coordinate system for this solid. If we do that here are the limits for the ranges.

0 6 θ 6 2π, 0 6 r 6 1, 0 6 z 6 4− 3r2.

We will also need the divergence of the vector field so let us get that.

div F =∂

∂x(xy) +

∂

∂y

„−1

2y2

«+

∂

∂z(z) = y − y + 1 = 1.

It follows from Divergence Theorem that the integral is

ZZ

S

F · n dσ =

Z 2π

0

Z 1

0

Z 4−3r2

0

r dz dr dθ

=

Z 2π

0

Z 1

0

`4r − 3r3´ dr dθ

=

Z 2π

0

»2r2 − 3

4r4

–1

0

dθ =

Z 2π

0

„5

4

«dθ =

5π

2.

2

222

5.6 Stokes’ Theorem


In this section we are going to discuss a generalization of Green’s Theorem to three dimensions. In Green’sTheorem we relate a line integral to a double integral over some region. In this section we will relate a lineintegral to a surface integral. However, before we give the theorem we first need to define the curve that weare going to use in the line integral.

Let us start with the following oriented surface with the indicated orientation.

x

y

z

Curve C

n n

n n

Surface S

If S is an oriented surface in 3-space bounded by a simple closed parametric curve C, then there are twopossible relationships between the orientations of S and C, which can be described as follows. Aroundthe edge of this surface we have a curve C. This curve is called the boundary curve. The orientationof the surface S will induce the positive orientation of C. To get the positive orientation of C thinkof yourself as walking along the curve. While you are walking along the curve if your head is pointing inthe same direction as the unit normal vectors while the surface is on the left then you are walking in thepositive direction on C.

The positive orientation of C establishes a right-hand relationship between the orientations of S andC in the sense that if the fingers of the right hand are curled from the direction of C towards S, thenthe thumb points (roughly) in the direction of the orientation of S.

In Section 5.3 (page 197) we defined the curl of a vector field F but we did not give a clear physicalexplanation of its meaning at that time. The following result, known as Stokes’ Theorem, will provideus with a physical interpretation of the curl in the context of fluid flow. Although the details of Physics isbeyond the scope of this text, however, a deeper understanding would still be helpful in applications.

Theorem 5.6.1 (Stokes’ Theorem) Let S be an oriented smooth surface that is bounded by asimple, closed, piecewise smooth boundary curve C with positive orientation. Also let F be a vectorfield then

I

C

F · dr =

ZZ

S

curl F · n dσ.

223


Recall that if F is a force field, the integral on the left side represents the work performed by the forcefield on a particle that traverses the curve C. Thus, loosely speaking, Stokes’ Theorem states:

The work performed by a force field on a particle that traverses a simple, closed, piecewise smoothcurve C in the positive direction can be obtained by integrating the normal component of the curlover an oriented surface S bounded by C.

For computational purposes Stokes’ Theorem is usually the method of choice for calculating work aroundpiecewise smooth curves with multiple sections, since it eliminates the need for a separate integral evaluationover each section. This is illustrated in the following example.

¥ Example 5.6.1 (Stokes’ theorem)

Find the work performed by the force field

F = z2 i + y2 j + xk

on a particle that traverses the triangle C with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1).

Solution The work performed by the field is

W =

I

C

F · dr.

However, to evaluate this line integral directly would require three separate integrations, one over each sideof the triangle. Instead, we will use Stokes’ Theorem to express the work as a surface integral

W =

ZZ

S

curl F · n dσ

in which the plane surface S enclosed by C is assigned an upwards orientation to make the orientationof C positive.

x

y

z

1

1

1

Plane Surface Sx + y + z = 1

n

Boundary curve C

224


We need the curl of the vector field eventually let us get that first.

curl F =

˛̨˛̨˛̨˛̨˛

i j k

∂

∂x

∂

∂y

∂

∂z

z2 y2 x

˛̨˛̨˛̨˛̨˛

= (2z − 1) j.

Now, all we have is the boundary curve for the surface that we will need to use in the surface integral.However, as noted above all we need is any surface that has this as its boundary curve. So, let us use thetriangular plane with upwards orientation for the surface.

Since the plane is oriented upwards this induces the positive direction on C as shown. The equationof this plane is

x + y + z = 1 =⇒ z = g(x, y) = 1− x− y.

Now, let us use Stokes’ Theorem and get the surface integral set up.

Z

C

F · dr =

ZZ

S

curl F · n dσ

=

ZZ

S

(2z − 1) j · n dσ

=

ZZ

R

(2z − 1) j · ∇f dA.

Now we need to find a couple of things. First let us get the gradient. Recall that this comes from thefunction of the surface.

f(x, y, z) := z − g(x, y) = z − 1 + x + y,

∇f = i + j + k.

Note as well that this also points upwards and so we have the correct direction.

Now, R is the region in the xy-plane shown below.

0x

y

1

1

y = −x + 1

Region R

��

��

225


We get the equation of the line by plugging in z = 0 into the equation of the plane. So based on this theranges that define R are

0 6 x 6 1, 0 6 y 6 −x + 1.

The integral is then

Z

C

F · dr =

ZZ

R

(2z − 1) j · (i + j + k) dA

=

Z 1

0

Z −x+1

0

[2(1− x− y)− 1] dy dx.

Do not forget to plug in for z since we are doing the surface integral on the plane. Finishing all gives

W =

Z

C

F · dr =

Z 1

0

Z −x+1

0

(1− 2x− 2y) dy dx

=

Z 1

0

hy − 2xy − y2

i−x+1

0dx

=

Z 1

0

`x2 − x

´dx

=

»1

3x3 − 1

2x2

–1

0

= −1

6.

2

In Stokes’ Theorem note that the surface S can actually be any surface so long as its boundary curveis given by C. This is something that can be used to our advantage to simplify the surface integral onoccasion. Let us take a look at an example.

¥ Example 5.6.2 (Stokes’ theorem)

Use Stokes’ Theorem to evaluateZZ

S

curl F · n dσ,

where F = z2 i− 3xy j + x3y3 k and S is the part of z = 5− x2 − y2 above the plane z = 1. Assumethat S is oriented upwards.

Solution As guaranteed by Stokes’ Theorem, the value of the surface integral is the same as the valueof the line integral

ZZ

S

curl F · n dσ =

I

C

F · dr.

Note that the line integral is simpler to evaluate and hence will be the method of choice in this case.

226


Let us start with the sketch of the surface.

x

y

z

5

1

n

Surface S

Curve C

In this case the boundary curve C will be where the surface intersects the plane z = 1 and so will be thecurve 1 = z = 5− x2 − y2, or

x2 + y2 = 4, at z = 1.

So, the boundary curve will be the circle of radius 2 that is in the plane z = 1. The parametrization ofthis curve is

r(t) = 2 cos t i + 2 sin t j + k, 0 6 t 6 2π.

The first two components give the circle and the third component makes sure that it is in the plane z = 1.

Using the Stokes’ Theorem we can write the surface integral as the following line integral.

ZZ

S

curl F · n dσ =

Z

C

F · dr =

Z 2π

0

F“r(t)

” · r′(t) dt.

So we need a couple of quantities before we do this integral. Let us first get the vector field evaluated on thecurve. Remember that this is simply plugging the components of the parametrization into the vector field.

F“r(t)

”= (1)2 i− 3(2 cos t)(2 sin t) j + (2 cos t)3(2 sin t)3 k

= i− 12 cos t sin t j + 64 cos3 t sin3 t k.

Next, we need the derivative of the parametrization and the dot product of this and the vector field.

r′(t) = −2 sin t i + 2 cos t j,

F“r(t)

” · r′(t) = −2 sin t− 24 cos2 t sin t.

227


The surface integral becomes

ZZ

S

curl F · n dσ =

Z 2π

0

`−2 sin t− 24 cos2 t sin t´

dt

=h2 cos t + 8 cos3 t

i2π

0= 0.

2

In both of these examples we are able to do an integral that could be somewhat unpleasant to deal with andby the use of Stokes’ Theorem we are able to convert it into an integral that is easier to evaluate.

Before we close this section let us give a relationship between Green’s Theorem and Stokes’ Theorem.

[Relationship between Green’s Theorem and Stokes’ Theorem]

It is sometimes convenient to regard a vector field

F(x, y) = P (x, y) i + Q(x, y) j

in 2-space as a vector field in 3-space by expressing it as

F(x, y) = P (x, y) i + Q(x, y) j + 0k.

If D is a region in the xy-plane enclosed by a simple, closed, piecewise smooth curve C, then we cantreat D a flat surface, and we can treat a surface integral over D as an ordinary double integral overD. Thus, if we orient D up and C counter-clockwise looking down the positive z-axis, then Stokes’Theorem applies to the above three-dimensional vector field and gives

I

C

F · dr =

ZZ

D

curl F · k dA.

But

curl F =

˛̨˛̨˛̨˛̨˛

i j k

∂

∂x

∂

∂y

∂

∂z

P Q 0

˛̨˛̨˛̨˛̨˛

=

„−∂Q

∂z

«i +

„∂P

∂z

«j +

„∂Q

∂x− ∂P

∂y

«k

since ∂Q/∂z = ∂P/∂z = 0. Substituting this expression into the double integral above and expressingthe integrals in terms of components yields

I

C

P dx + Q dy =

ZZ

D

„∂Q

∂x− ∂P

∂y

«dA

which is Green’s Theorem (page 190). Thus, we have shown that

Green’s Theorem can be viewed as a special case of Stokes’ Theorem.

228

notes100-ch5.pdf

Documents

Transcript of notes100-ch5.pdf