Integrability 7

Chapter 6

Integrability

Outline:

• The Riemann integral

• The line integral

• The line integral with respect to the arc length

• The double integral

6.1 Riemann integrable functions

Outline7.1 Divisions of a compact interval.7.2 The Riemann integral.7.3 Characterizations of Riemann integrability using sequences of divisions. Cauchy’s criterion.7.4 Darboux’s criterion on Riemann integrability.7.5 Classes of Riemann integrable functions.7.6 Lebesque’s criterion on Riemann integrability.7.7 Operations with Riemann integrable functions.7.8 Monotony properties of the Riemann integrable function.7.9 Mean value theorems7.10 The additivity of the integral function on the interval.7.11 Primitives, the primitivability of continuous functions, the Leibniz-Newton formula.7.12 Computing methods for the primitives.7.13 The integrability of rational functions.7.14 Methods of integration.

60

CHAPTER 6. INTEGRABILITY 61

6.2 The line integral

The line integral is a natural generalization of the definite Riemann integralR ba f (x) dx. In the integralR b

a f (x) dx we integrate the function f (x) from x = a to x = b along the line segment [a, b] from a tob. Replacing the line segment [a, b] by a general curve C in the plane or in the space, we obtain theline (or curve) integral

RC F · dr defined below.

We begin by defining the notion of curve (either in the plane R2 or in the space R3).

Definition 6.2.1 A curve C is the oriented image (graph) of a continuous function r : [a, b] → R3,r (t) = (x (t) , y (t) , z (t)), t ∈ [a, b] (see Figure 6.1).

The function r = r (t) is called a parametrization of the curve C, and r (a) is called the ini-tial/starting point of the curve C and r (b) is called the terminal/endpoint of the curve C. Ifr (a) = r (b) be say that the curve C is a closed curve.

If r is has a derivative r0 (t) = (x0 (t) , y0 (t) , z0 (t)) which is continuous at all points t ∈ (a, b) andr0 (t) 6= (0, 0, 0), we say that C is a smooth curve.

x

z

y

r(t) = (x(t), y(t), z(t))

r(b)

r(a)C

Figure 6.1: A curve C and its parametrization r (t) = (x (t) , y (t) , z (t)), t ∈ [a, b].

Remark 6.2.2 Some useful examples of curves/parametrizations are given below:

1. A parametrization of the line segment from A (x1, y1, z1) to B (x2, y2, z2) is given by r : [0, 1]→R3, where

r (t) = (1− t)A+ tB

= (1− t) (x1, y1, z1) + t (x2, y2, z2)

= ((1− t)x1 + tx2, (1− t) y1 + ty2, (1− t) z1 + tz2) , t ∈ [0, 1] .

2. A parametrization of an arc C of a circle of center (0, 0) and radius 1, between angles α and βis given by r : [α, β]→ R2, where

r (t) = (cos t, sin t) , t ∈ [α, β] .

More generally, if the circle has radius R and center (x0, y0), the corresponding parametrizationis

r (t) = (x0 +R cos t, y0 +R sin t) , t ∈ [α, β] .

Mihai N. Pascu — Mathematical Analysis lecture notes


3. A parametrization which is very useful in practice (in exercises) is that of the graph of a givencontinuous function f : [a, b]→ R2. In this case, the parametrization is given by r : [a, b]→ R2,where

r (t) = (t, f (t)) , t ∈ [a, b] .

For example, the parametrization of a part of a parabola, given by the graph of the functionf : [−2, 3]→ R2, f (x) = x2 − 5x+ 7 is given by

r (t) =¡t, t2 − 5t+ t

¢, t ∈ [−2, 3]

(to remember this easily, just set x = t ∈ [−2, 3] - the parameter, and then y = f (x) = f (t) =t2 − 5t+ 7, so r (t) = (x (t) , y (t)) =

¡t, t2 − 5t+ t

¢).

4. If C : r : [a, b] → R3 is a given curve, then the reversed curve C− obtaining by reversing theinitial and starting points of the curve C, has the parametrization r− : [a, b]→ R3, where

r− (t) = r (a+ b− t) , t ∈ [a, b] .

With this preparation, we can now give the definition of the line integral, as follows:

Definition 6.2.3 Given a continuous function F = (F1, F2, F3) defined on smooth curve C : r =r (t) = (x (t) , y (t) , z (t)), t ∈ [a, b], we define the line integral of F over the curve C byZ

CF · dr =

Z b

aF (r (t)) · r0 (t) dt

Remark 6.2.4 Given two vectors u = (u1, u2, u3) and v = (v1, v2, v3) in R3, u · v denotes the scalarproduct / dot product of the vectors u and v, defined by

u · v = u1v1 + u2v2 + u3v3.

The line integralRC F · dr defined above is therefore equal toZ

CF · dr =

ZC(F1, F2, F3) · (dx, dy, dz)

=

ZCF1dx+ F2dy + F3dz

=

Z b

a(F1 (x (t) , y (t) , z (t)) , F2 (x (t) , y (t) , z (t)) , F3 (x (t) , y (t) , z (t))) ·

¡x0 (t) , y0 (t) , z0 (t)

¢dt

=

Z b

aF1 (x (t) , y (t) , z (t)) x0 (t) + F2 (x (t) , y (t) , z (t)) y0 (t) + F3 (x (t) , y (t) , z (t)) z0 (t) dt.

This formula says that in order to compute the line integralRC F · dr we integrate the dot product

of the vectors F (r (t)) = (F1 (r (t)) , F2 (r (t)) , F3 (r (t))) and r0 (t) = (x0 (t) , y0 (t) , z0 (t)) over theinterval [a, b].

Remark 6.2.5 It is not immediate from the definition that the line integralRC F ·dr is independent

of the parametrization. That is, if r1 : [a, b]→ R3 and r2 : [c, d]→ R3 are two parametrizations ofthe same oriented curve C, thenZ b

aF (r1 (t)) · r01 (t) dt =

Z d

cF (r2 (t)) · r02 (t) dt.

However, this follows by the change of variables in the definite integral, as follows: if r1 and r2both describe the same smooth curve C, then it can be shown that r1 (t) = r2 (u (t)), for some bijective



function u : [a, b]→ [c, d]. With the change of variables t = u (s), we obtainZ c

dF (r2 (t)) · r02 (t) dt =

Z b=u−1(c)

a=u−1(d)F (r2 (u (s))) · r02 (u (s))u0 (s) ds

=

Z b

aF (r1 (s)) · r01 (s) ds

=

Z b

aF (r1 (t)) · r01 (t) dt.

Example 6.2.6 Compute the line integralRC F · dr where F (x, y, z) =

¡x, y2, sin z

¢and C is the arc

of the circle of center (0, 0) and radius 1 located in the first quadrant (i.e. x, y ≥ 0).A parametrization of the given curve is r : [0, π/2]→ R3, where

r (t) = (cos t, sin t, 0) , t ∈h0,π

2

i.

We have r0 (t) =¡(cos t)0 , (sin t)0 , (0)0

¢= (− sin t, cos t, 0), and therefore the given integral isZ

CF · dr =

Z π/2

0F (r (t)) · r0 (t) dt

=

Z π/2

0

¡cos t, sin2 t, sin 0

¢· (− sin t, cos t, 0) dt

=

Z π/2

0− sin t cos t+ sin2 t cos tdt

= −sin2 t

2+sin3 t

3

¯t=π/2t=0

=

µ−12+1

3

¶−µ−02+0

3

¶= −1

6.

Remark 6.2.7 (Physical interpretation of the line integral) In Physics, the work done by aconstant force acting on a particle moving in a straight line, is defined as W = Force ·distance. Ingeneral, given a force F (not necessarily constant) and a particle moving on a curve C (not necessarilya line segment), we can compute the work done by dividing the curve C by points r (tk) into smallparts Ck, k = 1, . . . , n, such that each curve Ck is approximately a straight line segment, and the forceF is approximately constant on Ck (see Figure 6.2).

It follows that the work done Wk in moving the particle along the curve Ck is approximately

Wk = F (r (tk)) · (r (tk)− r (tk−1)) ≈ F (r (tk)) · r0 (tk)∆tk

(note that the tangent vector r0 (tk) is a good approximation of the curve Ck by a straight line if thepoints r (tk) are sufficiently close to each other and the curve C is smooth), and therefore we obtain

W ≈W1 + . . .+Wn =nX

k=1

Wk =nX

k=1

F (r (tk)) · r (tk)∆tk.

To make this formula exact, we pass to the limit with n→∞, and we obtain

W = limn→∞

nXk=1

F (r (tk)) · r (tk)∆tk =Z b

aF (r (t)) · r0 (t) dt,

which shows that the line integralRC F ·dr gives the work done by a force F in moving a particle along

the curve C. This is the reason for which the line integralRC F · dr is sometimes also called the work

integral.



x

z

y

r(tk)

r(b)

r(a)

C

r(tk−1)

Ck

Figure 6.2: Physical interpretation of the line integralRC F · dr: work done by F in moving an object

along C.

Since the line integral is defined by a definite (Riemann) integral, it also has the correspondingproperties. More precisely, we have:

Theorem 6.2.8 (Properties of the line integral) If F , G are continuous functions defined on thesmooth curve C, then

1.RC

¡F + G

¢· dr =

RC F · dr +

RC G · dr (additivity)

2.RC

¡kF¢· dr = k

RC F · dr, for any constant k ∈ R (homogeneity)

3. If C = C1 ∪ C2 is formed by union of two curves, thenRC1∪C2 F · dr =

RC1

F · dr +RC2

F · dr(additivity with respect to the curve)

4.RC− F · dr = −

RC F · dr

Proof. Follow from the corresponding properties of the definite integral: the properties 1) and 2)follow from the properties Z b

a(f (t) + g (t)) dt =

Z b

af (t) dt+

Z b

ag (t) dt

and Z b

akf (t) dt = k

Z b

af (t) dt,

of the definite integral.The property 3) follows from the propertyZ b

af (t) dt =

Z c

af (t) +

Z b

cf (t) dt,

for any a < c < b, and the property 4) follows from the propertyZ b

af (t) dt = −

Z a

bf (t) dt.

For the definite integral, we saw (the Leibniz-Newton formula) that if F 0 (x) = f (x) for x ∈ [a, b],then Z b

af (x) dx = F (b)− F (a) .

We will see that with the appropriate definitions, a similar formula holds for the line integral.First, we introduce the concept of independence of path of a line integral, as follows:



Definition 6.2.9 We say that the line integralRC F · dr is independent of the path C of inte-

gration in the domain D, if for any two points A,B ∈ D and any two curves C1 and C2 containedin D with starting point A and endpoint B, the corresponding line integrals are equal:Z

C1

F · dr =ZC2

F · dr.

We have:

Theorem 6.2.10 If F = (F1, F2, F3) is continuous in a domain D ⊂ R3, then the line integralRC F · dr is independent of path in D if and only if there exists a function f : D→ R such that

F = grad f =

µ∂f

∂x,∂f

∂y,∂f

∂z

¶in D.

In this case, the line integralRC F · dr can be computed as follows:Z

CF · dr = f (B)− f (A) ,

where B is the endpoint and A is the starting point of the curve C.

Proof. Assume that there exists the function f : D → R such that F = grad f , that is F1 =∂f∂x ,

F2 =∂f∂y and F3 =

∂fdz .

Let C : r = r (t) = (x (t) , y (t) , z (t)) : [a, b] → R3 be a curve with starting point A = r (a) andendpoint B = r (b). and If the curve We haveZ

CF · dr =

Z b

aF (r (t)) · r0 (t) dt

=

Z b

aF1 (r (t))x

0 (t) + F2 (r (t)) y0 (t) + F3 (r (t)) z

0 (t) dt

=

Z b

a

∂f

∂x(r (t))x0 (t) +

∂f

∂y(r (t)) y0 (t) +

∂f

∂z(r (t)) z0 (t) dt

=

Z b

a

d

dtf (x (t) , y (t) , z (t)) dt

= f (r (t))|t=bt=a

= f (r (b))− f (r (a))

= f (B)− f (A) ,

and therefore the line integralRC F · dr does not depend on the path of integration (depends only on

its endpoint and starting point).Conversely, if the line integral

RC F ·dr does not depend on the path of integration, it can be shown

that the function f : D→ R defined by

f (x, y, z) =

ZCF · dr,

where C is an arbitrary curve contained in D with starting point (x0, y0, z0) ∈ D arbitrarily fixed andendpoint (x, y, z) ∈ D, is differentiable in D and satisfies the condition F = grad f .

To see this, we choose a curve C = C1 ∪ C2 such that the last part of the curve (the part close tothe endpoint (x, y, z)) is a line segment C2 = [(x− h, y, z) , (x, y, z)] parallel to the x - axis (see Figure6.3). We obtain:

f (x, y, z) =

ZCF · dr

=

ZC1

F · dr +ZC2

F · dr

= f (x− h, y, z) +

ZC2

F · dr,



Figure 6.3: The curve C = C1 ∪C2 from (x0, y0, z0) to (x, y, z).

and therefore

f (x, y, z)− f (x− h, y, z) =

ZC2

F · dr.

Using the parametrization C2 : r : [0, 1]→ R3 given by

r (t) = (1− t) (x− h, y, z) + t (x, y, z)

= (x− th, y, z) ,

we obtain

f (x, y, z)− f (x− h, y, z) =

ZC2

F · dr

=

Z 1

0F1 (x+ ht− h, y, z) · (x+ ht− h)0 + F2 (x+ ht− h, y, z) (y)0 +

F1 (x+ ht− h, y, z) (z)0 dt0 + F2 (x+ ht− h, y, z) · (y)0 + F1 (x+ ht− h, y, z) · (z)0 dt

=

Z 1

0F1 (x+ ht− h, y, z) · h+ F2 (x+ ht− h, y, z) · 0 + F1 (x+ ht− h, y, z) · 0dt

= h

Z 1

0F1 (x+ ht− h, y, z) dt,

and therefore

limh→0

f (x, y, z)− f (x− h, y, z)

x− (x− h)= lim

h→0

Z 1

0F1 (x+ ht− h, y, z) dt

= F1 (x, y, z) ,

which shows that f has partial derivative with respect to x

∂f

∂x(x, y, z) = F1 (x, y, z) .

A similar proof shows that ∂f∂y = F2 and

∂f∂z = F3, and therefore F = (F1, F2, F3) =

³∂f∂x ,

∂f∂y ,

∂f∂z

´=

grad f , concluding the proof.

Example 6.2.11 Consider the line integralRC F ·dr, where F = (x+ y, y + x, cos z) and C is a curve

from A = (0, 0, 0) to B = (1, 2, π/2) .



It can be checked that F = grad³x2

2 +y2

2 + sin z´, and therefore from the above theorem it follows

that ZCF · dr =

x2

2+

y2

2+ sin z

¯(1,2,π/2)(0,0,0)

=

µ1

2+22

2+ sin

π

2

¶−µ0

2+ 02 + sin 0

¶=

1

2+ 2 + 1

=7

2.

In general, it is not very easy to see whether a given function F is the gradient of a certain functionf . The following theorem gives us a simple criterion for deciding whether F = grad f :

Theorem 6.2.12 Let D be a simply connected domain and F = (F1, F2, F3) have continuous partialderivatives. Then

RC F · dr is independent of path in D if and only if curl F = 0, where

curl F =

¯¯ i j k

∂∂x

∂∂y

∂∂z

F1 F2 F3

¯¯ = µ∂F3∂y

− ∂F2∂z

,−∂F3∂x

+∂F1∂z

,∂F2∂x− ∂F1

∂y

¶.

Proof. AssumeRC F ·dr is independent of path. By the previous theorem, F = grad f , or F1 = ∂f

∂x ,F2 =

∂f∂y and F3 =

∂f∂z for some differentiable function f : D→ R.

We have

∂F3∂y− ∂F2

∂z=

∂

∂y

µ∂f

∂z

¶− ∂

∂z

µ∂f

∂y

¶=

∂2f

∂y∂z− ∂2f

∂z∂y= 0,

by Schwarz’s theorem (since F has continuous partial derivatives and F = grad f =³∂f∂x ,

∂f∂y ,

∂f∂z

´, f

has continuous second order partial derivatives, and therefore ∂2f∂y∂z =

∂2f∂z∂y ).

Similar proof shows −∂F3∂x +

∂F1∂z = 0 and

∂F2∂x −

∂F1∂y = 0, and therefore curl F = (0, 0, 0) = 0.

We will not give here the proof of the converse (it requires Stokes’s theorem).



6.3 The line integral with respect to the arc length

We will first introduce the notion of arc length of a curve. Consider C a smooth curve, with para-metrization r : [a, b] → R3, r (t) = (x (t) , y (t) , z (t)). To approximate the length of the curve C, weconsider a partition ∆ : a = t0 < t1 < . . . < tn = b and we approximate the length of the curveC by the length of the polygonal line passing through the points Pi = r (ti) = (x (ti) , y (ti) , z (ti)),i = 0, 1, . . . , n. We have

Length (C) ≈nXi=1

||r (ti)− r (ti−1)|| .

Applying Lagrange theorem to the function r : [ti−1, ti] → R it follows that there exists a pointt∗i ∈ [ti−1, ti] such that r (ti)− r (ti−1) = r0 (t∗i ) (ti − ti−1), therefore we obtain:

Length (C) ≈nXi=1

||r (ti)− r (ti−1)||

=nXi=1

¯¯r0 (t∗i ) (ti − ti−1)

¯¯=

nXi=1

q(x0 (t∗i ))

2 + (y0 (t∗i ))2 + (z0 (t∗i ))

2 (ti − ti−1)

=nXi=1

q(x0 (t∗i ))

2 + (y0 (t∗i ))2 + (z0 (t∗i ))

2∆ti

To make the above approximation exact, we pass to the limit with ||∆|| = max1≤i≤n

(ti − ti−1) → 0.

Since C is a smooth curve, by the definition of the Riemann integral it follows that the limit of theabove sum exists and we have

lim||∆||&0

nXi=1

q(x0 (t∗i ))

2 + (y0 (t∗i ))2 + (z0 (t∗i ))

2∆t =

Z b

a

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt

We are thus lead to the following:

Definition 6.3.1 Given a smooth curve C with parametrization r : [a, b]→ R, r (t) = (x (t) , y (t) , z (t)),we define the length of the curve C by

Length (C) =

Z b

a

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt.

Remark 6.3.2 In the definition above, the formula which gives the length of the curve C depends onthe parametrization r = r (t). Since there are many possible parametrizations of a given smooth curveC, we should check to see if the above definition is independent on the choice of the parametrization.However, this follows by the independence of the line integral with respect to the arc length (see Remark6.3.5 below).

Example 6.3.3 Let’s use the above formula to compute the length of the circle C of radius R centeredat the origin.

A parametrization of C is given by r : [0, 2π]→ R, r (t) = (R cos t, R sin t, 0).We have r0 (t) = (−R sin t, R cos t, 0), and using the definition above we obtain

Length (C) =

Z 2π

0

q(−R sin t)2 + (R cos t)2 + 02dt

=

Z 2π

0Rdt

= Rt|2π0= 2πR,



which coincides with the formula known from geometry (i.e. the length of a circle of radius R is equalto 2πR).

To make it easier to remember the next definition, we consider a small part of the curve C in theprevious discussion, say between the points r (t) and r (t+∆t). By the above definition, its length∆s is given by

∆s =

Z t+∆t

t

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt,

and by the mean value theorem we obtain

∆s =

q(x0 (t∗))2 + (y0 (t∗))2 + (z0 (t∗))2∆t,

where t∗ is a intermediate point between t and t+∆t.Passing to the limit with ∆t→ 0, it follows that

lim∆t→0

∆s

∆t= lim

∆t→0

q(x0 (t∗))2 + (y0 (t∗))2 + (z0 (t∗))2 =

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2,

ords

dt=

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2.

This formula shows that the length s = s (t) of an infinitesimal arc of the curve C (called the arclength of the curve C) has length

ds =

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt =

¯¯r0 (t)

¯¯dt,

and justifies the following definition:

Definition 6.3.4 Given a continuous function f = f (x, y, z) defined on a smooth curve C : r =r (t) = (x (t) , y (t) , z (t)), t ∈ [a, b], we define the line integral of f with respect to the arc length of thecurve C byZ

Cfds =

Z b

af (r (t))

¯¯r0 (t)

¯¯dt =

Z b

af (r (t))

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt

Remark 6.3.5 It is not immediate from the definition that the line integralRC fds is independent

of the parametrization. That is, if r1 : [a, b]→ R3 and r2 : [c, d]→ R3 are two parametrizations ofthe same oriented curve C, thenZ b

af (r1 (t))

¯¯r01 (t)

¯¯dt =

Z d

cf (r2 (t))

¯¯r02 (t)

¯¯dt.

However, this follows by the change of variables in the definite integral, as follows: if r1 and r2both describe the same smooth curve C, then it can be shown that r1 (t) = r2 (u (t)), for some bijectivefunction u : [a, b]→ [c, d].

With the change of variables t = u (s), we obtainZ d

c(r2 (t))

¯¯r02 (t)

¯¯dt =

Z u−1(d)

u−1(c)f (r2 (u (s)))

¯¯r02 (u (s))

¯¯u0 (s) ds

=

Z b

af (r2 (u (s)))

¯¯r02 (u (s))u

0 (s)¯¯ds

=

Z b

af (r1 (s))

¯¯(r2 (u (s)))

0¯¯ ds=

Z b

af (r1 (s)) ·

¯¯r01 (s)

¯¯ds.



Example 6.3.6 Compute the line integralRC x+ 2y − zds where C is the arc of the circle of radius

1 centered at (0, 0) located in the first quadrant (i.e. x, y ≥ 0) oriented counterclockwise.A parametrization of the given curve is r : [0, π/2]→ R3, where

r (t) = (cos t, sin t, 0) , t ∈h0,π

2

i.

We have r0 (t) =¡(cos t)0 , (sin t)0 , (0)0

¢= (− sin t, cos t, 0), and therefore the given integral isZ

Cx+ 2y − zds =

Z π/2

0(cos t+ 2 sin t− 0)

q(− sin t)2 + (cos t)2 + 02dt

=

Z π/2

0cos t+ 2 sin tdt

= sin t− 2 cos t|t=π/2t=0

= (1− 2 · 0)− (0− 2 · 1)= 3.

Since the line integral with respect to the arc length is defined by a definite (Riemann) integral, italso has the corresponding properties. More precisely, we have:

Theorem 6.3.7 (Properties of the line integral wrt arc length) If f, g are continuous func-tions defined on the smooth curve C, then

1.RC (f + g) ds =

RC fds+

RC gds (additivity)

2.RC (kf) ds = k

RC fds, for any constant k ∈ R (homogeneity)

3. If C = C1∪C2 is formed by union of two curves, thenRC1∪C2 fds =

RC1

fds+RC2

fds (additivitywith respect to the curve)

4.RC− fds =

RC fds (independence on orientation)

Proof. Consider r : [a, b]→ R, r (t) = (x (t) , y (t) , z (t)) a parametrization of the curve C.By definition, we haveZ

C(f + g) ds =

Z b

a(f (r (t)) + g (r (t)))

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt

=

Z b

af (r (t))

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt+Z b

ag (r (t))

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt

=

ZCfds+

ZCgds,

and similarly ZC(kf) ds =

Z b

a(kf (r (t)))

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt

= k

Z b

af (r (t))

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt+

= k

ZCfds,

using the additivity and homogeneity of the Riemann integral.To prove 3), since C = C1∪C2, there exists a point c ∈ (a, b) such that r1 : [a, c]→ R, r1 (t) = r (t)

is a parametrization of C1 and r2 : [c, b]→ R, r2 (t) = r (t) is a parametrization of C2.



From the definition of the line integral with respect to the arc length, we obtainZC1∪C2

fds =

Z b

af (r (t))

¯¯r0 (t)

¯¯dt

=

Z c

af (r (t))

¯¯r0 (t)

¯¯dt+

Z b

cf (r (t))

¯¯r0 (t)

¯¯dt

=

Z c

af (r1 (t))

¯¯r01 (t)

¯¯dt+

Z b

cf (r2 (t))

¯¯r02 (t)

¯¯dt

=

ZC1

fds+

ZC2

fds.

To prove part 4), note that a parametrization for the reversed curve C− is r1 : [a, b] → R,r1 (t) = r (t), and that we have r

01 (t) =

ddt (r (a+ b− t)) = −r0 (a+ b− t). Using the definition and

using the substitution u = a+ b− t (note that du = −dt), we obtainZC−

fds =

Z b

af (r1 (t))

¯¯r01 (t)

¯¯dt

=

Z b

af (r (a+ b− t)) ||−r (a+ b− t)|| dt

= −Z a

bf (r (u)) ||r (u)|| du

=

Z b

af (r (u)) ||r (u)|| du

=

ZCfds,

which concludes the proof.There are 3 applications of the line integral with respect to the arc length, as follows:

1. Length of a curve

It is immediate from the two definitions of above that the length of a curve C can be computedusing the line integral:

Length (C) =

ZC1ds

(just compare the first definition above with the second one, in which f ≡ 1)

2. Mass of a curve C with mass density f = f (x, y, z)

The mass of a curve with variable mass density f (for example a wire which is not uniform,having parts that are heavier/lighter) can be computed using the line integral as follows:

Mass (C) =

ZCfds.

To see why this formula is so, divide the curve C into small parts by points r (ti), i = 1, . . . , n,where r : [a, b] → R is a parametrization of C and a = t0 < t1 < . . . < tn = b is a partition of[a, b].

If the partition is fine enough and the mass density is continuous, the part of the curve C betweenpoints r (ti−1) and r (ti) has approximately constant density, hence its mass is approximatelyf (r (ti)) (density) times the length of this arc. Summing up the masses of all such arcs, weobtain

Mass (C) ≈nXi=1

f (r (ti))

q(x0 (t∗i ))

2 + (y0 (t∗i ))2 + (z0 (t∗i ))

2dt.



To make the above approximation exact, we pass to the limit with the norm of the partitiontending to zero, and we obtain

Mass (C) =

Z b

af (r (t))

q(x0 (t))2 + (y0 (t))2 + (z0 (t))2dt,

which by definition equalsRC fds.

3. The center of mass G (xG, yG, zG) of a given curve with mass density f = f (x, y, z) is given by

xG =

RC xfdsRC fds

, yG =

RC yfdsRC fds

, zG =

RC zfdsRC fds

.

The reason for the above formulae comes from Physics (the sum of moments with respect to thecenter of mass of all forces involved equals zero), but we will not discuss it here.

Example 6.3.8 Find the length, mass and center of mass of the curve C with mass density f (x, y, z) =3x, where C is the arc of parabola y = x2 lying in the first quadrant.

A parametrization of the curve C is r : [0, 1]→ R, r (t) =¡t, t2, 0

¢.

From the above formulae we obtain:- Length of C

Length (C) =

ZC1ds

=

Z 1

0

q(1)2 + (2t)2 + 02dt

= 4

Z 1

0

r1

4+ t2dt

Using the formulaR √

a2 + x2dx = x2

√a2 + x2 + a2

2 ln³x+√a2 + x2

´+ C, we obtain

Length (C) = 4

Ãt

2

r1

4+ t2 +

1/4

2ln

Ãt+

r1

4+ t2

!!¯¯1

0

= 4

Ã1

2

√5

2+1

8ln

Ã1 +

√5

2

!!

=√5 +

1

2ln

Ã1 +

√5

2

!

- Mass of C

Mass (X) =

ZCfds

=

Z 1

03t

q(1)2 + (2t)2 + 02dt

= 3

Z 1

0tp1 + 4t2dt.



Using the substitution u = 1 + 4t2 (note that du = 8tdt) we obtain

Mass (X) = 3

Z 1

0tp1 + 4t2dt

= 3

Z 5

1

√u1

8du

=3

8

Z 5

1u1/2du

=3

8

u3/2

3/2

¯¯5

1

=5√5− 14

.

- Center of mass of C: G (xG, yG, zG)For example, the y coordinate of the center of mass is

yG =

RC yfdsRC fds

=

R 10 t

2 · 3t√1 + 4t2dt

5√5−14

=12

5√5− 1

Z 1

0t3p1 + 4t2dt

With the substitution u = 1 + 4t2 (note that du = 8tdt), we obtain

yG =12

5√5− 1

Z 5

1

u− 14

√u1

8du

=3

8¡5√5− 1

¢ Z 5

1u3/2 − u1/2du

=3

8¡5√5− 1

¢ Ãu5/2

5/2− u3/2

3/2

!¯¯5

1

=3

8¡5√5− 1

¢ Ã55/25/2− 5

3/2

3/2

!− 3

8¡5√5− 1

¢ µ 1

5/2− 1

3/2

¶=

15√5

8¡5√5− 1

¢ µ2− 23

¶− 3

8¡5√5− 1

¢ µ− 415

¶=

5√5

2¡5√5− 1

¢ + 1

10¡5√5− 1

¢=

25√5 + 1

10¡5√5− 1

¢ ,and similarly for xG and zG.



6.4 The double integral

The double integral can be viewed as an extension of the definite integralR ba f (x) dx, as follows.

Consider a function of two variables, f = f (x, y) : D → R defined on a domain D ⊂ R2. Wedefined the integral by dividing the interval [a, b] into n subintervals [xi−1, xi], i = 1, . . . , n, of length∆x = xi − xi−1 and taking limits of the sums of the form

nXi=1

f (x∗i )∆x,

where x∗i ∈ [xi−1, xi] are arbitrarily chosen points in the interval [xi−1, xi], i = 1, . . . , n.Proceeding similarly, we can divide the given region D into m · n rectangles with sides parallel to

the x and y axes (see Figure 6.4), and we can form the sumnXi=1

mXj=1

f¡x∗i , y

∗j

¢∆x∆y,

where³x∗i , y

∗j

´is an arbitrary point in the rectangle Rij , i = 1, . . . , n and j = 1, . . . ,m.

Figure 6.4: The domain D dividided into rectangles Rij .

If the function f is continuous on D and D is a “nice” domain (its boundary is a smooth curve),it can be shown that the limit as ∆x,∆y → 0 of the above sum exists and is finite, and we denote thelimit by Z Z

Df (x, y) dxdy = lim

∆x,∆y→0

nXi=1

mXj=1

f¡x∗i , y

∗j

¢∆x∆y, (6.1)

and and call it the double integral of f over the domain D.The double integral has similar properties with the definite integral, as follows:

Theorem 6.4.1 Let f, g : D→ R be continuous functions defined on the domain D ⊂ R2. We have:

1.R R

D (f (x, y) + g (x, y)) dxdy =R R

D f (x, y) dxdy +R R

D g (x, y) dxdy (additivity)

2.R R

D kf (x, y) dxdy = kR R

D f (x, y) dxdy for any constant k ∈ R (homogeneity)

3. If D = D1 ∪D2, thenR R

D1∪D2f (x, y) dxdy =

R RD1

f (x, y) dxdy +R R

D2f (x, y) dxdy

4. There exists (x0, y0) ∈ D such thatR R

D f (x, y) dxdy = f (x0, y0) ·Area (D)

6.4.1 Evaluation of the double integral

There are essentially two different ways for computing the double integral: as an iterate (succesive)integral or by a change of variables (see also Green’s theorem for an alternate possibility for theevaluation of the double integral).



Figure 6.5: The domain D between the graph of two functions g (x) ≤ y ≤ h (x), x ∈ [a, b].

6.4.2 Double integral as an iterate integral

Assume that the domain D is the region between the graphs of two functions g, h : [a, b] → R of thevariable x (see Figure 6.5), that is:

D = {(x, y) : g (x) ≤ y ≤ h (x) , x ∈ [a, b]} .

Computing the limit in (6.1) as an iterate integral, we obtainZ ZDf (x, y) dxdy = lim

∆x,∆y→0

nXi=1

mXj=1

f¡x∗i , y

∗j

¢∆x∆y

= lim∆x→0

nXi=1

⎛⎝ lim∆y→0

mXj=1

f¡x∗i , y

∗j

¢∆y

⎞⎠∆x=

Z b

a

ÃZ h(x)

g(x)f (x, y) dy

!dx,

and therefore in this case we can compute the double integral by first integrating the function ffirst with respect to y (between the limit g (x) and h (x)) and then integrating the resulting functionwith respect to x (between the limits a and b):Z Z

Df (x, y) dxdy =

Z b

a

ÃZ h(x)

g(x)f (x, y) dy

!dx. (6.2)

A similar formula holds if we reverse the roles of the variables x and y, that is if the domain Dcan be described as

D = {(x, y) : p (y) ≤ x ≤ q (y) , y ∈ [c, d]} ,

for some functions p, q : [c, d]→ R of the variable y (see Figure 6.6).In this case, we can compute the double integral by first integrating with respect to x (between

p (y) and q (y)) and then integrating the resulting function with respect to the variable y (between cand d): Z Z

Df (x, y) dxdy =

Z d

c

ÃZ q(y)

p(y)f (x, y) dx

!dy. (6.3)

Example 6.4.2 Compute the double integralR R

D ydxdy where D is the upper half of the unit diskx2 + y2 < 1.



Figure 6.6: The domain D between the graph of two functions p (y) ≤ x ≤ q (y), y ∈ [c, d].

The given domain D is of the form in Figure 6.5 (sketch a picture!), where g (x) = 0 and h (x) =√1− x2, x ∈ [−1, 1], and also of the form in Figure 6.5, where p (y) = −

p1− y2 and q (y) =

p1− y2,

y ∈ [0, 1].Using the first representation above, we haveZ Z

Dydxdy =

Z 1

−1

ÃZ √1−x2

0ydy

!dx

=

Z 1

−1

y2

2

¯y=√1−x2y=0

dx

=

Z 1

−1

1− x2

2dx

=x

2− x3

6

¯x=1x=−1

=

µ1

2− 16

¶−µ−12− −16

¶=

2

3.

Using the second representation above, we haveZ ZDydxdy =

Z 1

0

ÃZ √1−y2−√1−y2

ydx

!dy

=

Z 1

0xy|x=

√1−y2

x=−√1−y2

dy

=

Z 1

02yp1− y2dy

= −23

¡1− y2

¢3/2 ¯y=1y=0

=2

3,

obtaining the same result.



6.4.3 Change of variables in the double integral

Recall that for the definite integralR ba f (x) dx, using the change of variable x = x (u) : [α, β]→ [a, b],

we obtain Z b

af (x) dx =

Z β

αf (x (u))x0 (u) du.

Similarly, changing the variables in the double integralR R

D f (x, y) dxdy by½x = x (u, v)y = y (u, v)

, (u, v) ∈ D∗ , (6.4)

we obtain the following change of variables formulaZ ZDf (x, y) dxdy =

Z ZD∗

f (x (u, v) , y (u, v))

¯∂ (x, y)

∂ (u, v)

¯dudv. (6.5)

The formula says that changing the variables from (x, y) ∈ D → (u, v) ∈ D∗, we have to change

accordingly the domain of integration (from D to D∗), and the area element dxdy by¯∂(x,y)∂(u,v)

¯dudv,

where∂ (x, y)

∂ (u, v)=

¯∂x∂u

∂y∂u

∂x∂u

∂y∂v

¯is the Jacobian of the change of variables (6.4) - the determinant of the matrix of partial derivativesof the “old” variables x, y with respect to the “new” variables u, v (note that the corresponding“Jacobian” for the definite integral is just dx

du = x0 (u)).

Remark 6.4.3 To see why this formula is so, we go back to the definition (6.1), and we see that thearea element dxdy is the limiting area ∆x∆y of a rectangle, and similarly for dudv and ∆u∆v.

Approximating x = x (u, v) and y = y (u, v) by a Taylor’s formula of order 1 near the point (ui, vj),we obtain a linear transformation½

x (u, v) ≈ x (ui, vi) +∂x∂u (u− ui) +

∂x∂v (v − vj)

y (u, v) ≈ y (ui, vi) +∂y∂u (u− ui) +

∂y∂v (v − vj)

,

where all the partial derivatives are evaluated at the point (ui, vj).It can be shown that under a linear transformation, the area element element changes by a factor

equal to the absolute value of the Jacobian, that is

∆x∆y =

¯∂ (x, y)

∂ (u, v)

¯∆u∆v,

which by initial remark shows that

dxdy =

¯∂ (x, y)

∂ (u, v)

¯dudv.

Example 6.4.4 Compute the integralR R

D ex2+y2dxdy where D is the disk x2 + y2 < 9.

By using the change of variables (polar coordinates):½x = r cos ty = r sin t

, r ∈ [0, 3] and t ∈ [0, 2π] ,

the corresponding Jacobian is

∂ (x, y)

∂ (r, t)=

¯∂x∂r

∂y∂r

∂x∂t

∂y∂t

¯=

¯cos t sin t−r sin t r cos t

¯= r cos2 t+ r sin2 t = r,

and therefore we obtainZ ZDex

2+y2dxdy =

Z Z[0,3]×[0,2π]

er2 cos2 t+r2 sin2 trdrdt.



Computing the las integral as an iterated intgral, we haveZ ZDex

2+y2dxdy =

Z 2π

0

µZ 3

0rer

2dr

¶dt

=

Z 2π

0

1

2er

2

¯r=3r=0

dt

=

Z 2π

0

e9

2dt

=e9

2

¯t=2πt=0

= πe9.

6.4.4 Green’s Theorem

Green’s theorem gives us a formula which relates the double integral and the line integral, introducedin the previous section.

Theorem 6.4.5 (Green’s theorem) Let D ⊂ R2 be a bounded domain, having as boundary asmooth curve C (oriented such that the domain is on the left of the curve C). If F = (F1, F2) : D→ Ris continuous on D and F1,2 have continuous partial derivatives on D1, thenZ Z

D

µ∂F2∂x− ∂F1

∂y

¶dxdy =

ZC(F1, F2) · dr.

Remark 6.4.6 Note that the integrand on the left is just the third component of the curl F , whereF = (F1, F2, F3). More generally, a similar formula holds (Stokes’s theorem), which states thatZ Z

Scurl

¡F¢· ndA =

ZCF dr,

where S is an oriented surface with normal vector n and C is the boundary of the surface S, orientedaccordingly (such that the surface S is “to the left” of C).

Proof. It is sufficient to prove the theorem in the case when the domain D is of the form in Figure6.5 or in Figure 6.6 (we can reduce the proof to this case by dividing the domain D into domains ofthis form and then use the property 3) in Theorem 6.4.1).

Assume therefore that the domainD is of the form in Figure 6.5, that isD = {(x, y) : g (x) ≤ y ≤ h (x) , x ∈ [a,Computing the double integral as an itrated integral, we haveZ Z

D

∂F1∂y

(x, y) dxdy =

Z b

a

ÃZ h(x)

g(x)

∂F1∂y

(x, y) dy

!dx

=

Z b

aF1 (x, y)|y=h(x)y=g(x)

dx

=

Z b

aF1 (x, h (x))− F1 (x, g (x)) dx

=

Z b

aF1 (x, h (x)) dx−

Z b

aF1 (x, g (x)) dx.

The boundary curve C of D can be written as the union of two curves

C = Cg ∪ C−h ,1This means that F1,2 have partial derivatives in a domain D∗ containing D.



where Cg is the “lower” curve given by y = g (x), with parametrization

Cg : r (t) = (t, g (t)) , t ∈ [a, b],

and Ch is the “upper” curve given by y = h (x), with parametrization

Ch : r (t) = (t, h (t)) , t ∈ [a, b] .

We haveZC(F1, 0) · dr =

ZCg

(F1, 0) · dr +ZC−h

(F1, 0) · dr

=

ZCg

(F1, 0) · dr −ZCh

(F1, 0) · dr

=

Z b

a(F1 (t, g (t))) ·

¡1, g0 (t)

¢dt−

Z b

a(F1 (t, h (t))) ·

¡1, h0 (t)

¢dt

=

Z b

aF1 (t, g (t)) dt−

Z b

aF1 (t, h (t)) dt

= −Z Z

∂F1∂y

dxdy.

A similar proof shows that we also haveZC(0, F2) · dr =

Z Z∂F2∂x

dxdy,

and therefore adding the last two equalities, we obtainZC(F1, F2) · dr =

Z ZD

∂F2∂x− ∂F1

∂ydxdy,

concluding the proof.

6.4.5 Applications of the double integral

There are several applications of the double integral, among which we will reffer to the computationof volumes, areas and center of mass.

Given a continuous function z = f (x, y) : D → R with f ≥ 0 in D, the volume of the regionbeneath the graph of the function f and above the domain D (see Figure ??) is given by

Volume =

Z ZDf (x, y) dxdy.

Remark 6.4.7 To see why this is so, remember that the double integral is the limit of the sum in(6.1), in which f (xi, yj)∆x∆y gives an approximation of the parallelipiped with base the rectangleRij and height f (xi, yj). Summing up over i, j and taking the limit, this formula (that is, the doubleintegral) gives the area under the graph of the function z = f (x, y).

Example 6.4.8 We can compute the volume of a sphere of radius R (i.e. x2+y2+z2 < R2) by usingthe double integral as follows: the volume of the sphere is by symmery twice the volume under thegraph of the function f (x, y) =

pR2 − x2 − y2 above the domain D : x2+y2 < R2 (sketch a picture!).

Therefore, we obtain

Volume = 2

Z ZD

pR2 − x2 − y2dxdy.

To compute this integral, we can either use an iterate integral, or a substitution. Choosing thesubstitution (change to polar coordinates)½

x = r cos ty = r sin t

, r ∈ [0, R] and t ∈ [0, 2π] ,



the corresponding Jacobian is

∂ (x, y)

∂ (r, t)=

¯∂x∂r

∂y∂r

∂x∂t

∂y∂t

¯=

¯cos t sin t−r sin t r cos t

¯= r cos2 t+ r sin2 t = r,

and therefore we obtain

Volume = 2

Z Z[0,R]×[0,2π]

pR2 − r2 cos2 t− r2 sin2 trdrdt

= 2

Z 2π

0

µZ R

0rpR2 − r2dr

¶dt

= 2

Z 2π

0−13

¡R2 − r2

¢3/2 ¯r=Rr=0

dt

= 2

Z 2π

0

1

3

¡R2¢3/2

dt

=2

3R3t

¯t=2πt=0

=4

3πR3.

Taking f (x, y) ≡ 1 and remembering that a cylinder with a height of 1 has volume equal to thearea of its base, from the previous formula we obtain a formula for computing the area of a givendomain D:

Area (D) =

Z ZD1dxdy.

Example 6.4.9 We can compute the area of the disk D : x2 + y2 < R2 by using the double integralas follows:

Area =

Z ZD1dxdy.

Writing the domain D as the domain between the curves g (x) = −√R2 − x2 and h (x) =

√R2 − x2,

with x ∈ [−R,R], we can compute the double integral as an iterated integral, as folows:

Area =

Z R

−R

ÃZ √R2−x2

−√R2−x2

1dy

!dx

=

Z R

−R2pR2 − x2dx.

To compute the integral, we can either use the formulaR √

a2 − x2dx = x2

√a2 − x2+ a2

2 arcsinxa+C

or we can use a substitution, say x = R sin t, with t ∈ [−π/2, π/2], and after computation we obtain

Area = πR2.

Interpreting the function f (x, y) ≥ 0 as the density of mass at the point (x, y) ∈ D, and proceedingsimilarly with the previous remark, we see that the mass of a domain D with mass density f (x, y) isgiven by

Mass (D) =

Z ZDf (x, y) dxdy.

The center of mass of D can be computed by the following formulae:

xM =

R RD xf (x, y) dxdyR RD f (x, y) dxdy

and yM =

R RD yf (x, y) dxdyR RD f (x, y) dxdy

.



6.5 Exercises

1. Find the following antiderivatives:

a)Z6x2 + 8x+ 3dx b)

Z p2pxdx c)

Z14√xdx

d)Z

1√8− x2

dx e)Z

1

x2 + 7dx f)

Zdx

x2 − 10g)Z

x

(x− 1) (x+ 1)2dx h)

Zdx

x3 − 2x2 + xi)Z

dx

2x2 + 3x+ 2

j)Z

2x+ 3

x2 + 5x+ 6dx k)

Zx4

x3 − 1dx l)Z

1

x4 + 1dx

2. Find the following antiderivatives using a change of variables:

a)Z √

lnx

xdx b)

Zex

ex + 1dx c)

Zdx√

1− 25x2

d)Z

x3√1− x8

dx e)Z

cosxp2 + cos (2x)

dx f)Z

x√1 + x4

dx

g)Z

x¡1− x2

¢1 + x4

dx h)Z √

x√1− x3

dx i)Z p

x2 + x+ 1dx

3. Find the following antiderivatives using integration by parts:

a)Z

xe−xdx b)Z

xe2xdx c)Z

x sinxdx

d)Z

x lnxdx e)Zlnxdx f)

Zx2 ln (2x) dx

4. Evaluate the following definite integrals:

a)Z 1

0

1

1 + xdx b)

Z x

−xetdt c)

Z x

0cos tdt

d)Z 1

0

x

x2 + 3x+ 2dx e)

Z 4

1

1 +√x

x2dx f)

Z −1

−2

x

x2 + 4x+ 5dx

g)Z 1

0

x3

x8 + 1dx h)

Z π/3

π/6cotxdx i)

Z π/2

0sin (2x) dx

j)Z 1

0x2exdx k)

Z b

a

p(x− a) (x− b)dx l)

Z π

0x2 cosxdx

m)Z 1

0

x+ 1√1 + x2

dx n)Z 1

0

1

ex + e−xdx o)

Z 4

0xpx2 + 9dx

p)Z π/4

0ln (1 + tanx) dx q)

Z 1

0x2 arctanxdx r)

Z π

0x2 sin2 xdx

5.

6. Evaluate the line integralRC F · dr in the following cases:

(a) F (x, y, z) = (−y,−xy, 0) and C is the quarter of the circle of radius 1 in the first quadrant,traversed counterclockwise.

(b) F (x, y, z) = (z, x, y) and C is the helix given by r : [0, 2π]→ R, r (t) = (cos t, sin t, 3t).



(c) F (x, y) =¡y2,−x2

¢and C is the straight line segment from A = (0, 0) to B = (1, 4).

(d) F (x, y) =¡y2,−x2

¢and C is the graph of the function y = 4x2 from A = (0, 0) to

B = (1, 4).

7. Is the line integralRC F ·dr, where F (x, y) =

¡y2,−x2

¢independent on path in R2? (see Exercise

6c) and 6d) above)

8. Consider F (x, y, x) =¡5z, xy, x2z

¢and the curves C1 given by r1 (t) = (t, t, t) , t ∈ [0, 1] and C2

given by r2 (t) =¡t, t, t2

¢, t ∈ [0, 1].

(a) EvaluateRC1

F · dr(b) Evaluate

RC2

F · dr(c) Is the line integral

RC F · dr independent on path? (note that C1 and C2 have the same

initial and final point)

9. Show that the following line integrals are independent on path in R3. Determine a function fsuch that F = grad f and then compute the corresponding line integral

RC F · dr.

(a) F (x, y, z) = (3y, 3x, 2z), C is a curve from (0, 0, 0) to (4, 1, 2);

(b) F (x, y, z) = (ex cos y,−ex sin y, 0), C is a curve from (0, π, 1) to (3, π/2,−3);(c) F (x, y, z) =

¡3z2, 0, 6xz

¢, C is a curve from (−1, 0, 5) to (4, 7, 3);

(d) F (x, y, z) =³ex−y+z

2,−ex−y+z2 , 2zex−y+z2

´, C is a curve from (0,−1, 1) to (2, 4, 0);

(e) F (x, y, z) = (−z sin (xz) , cos y,−x sin (xz)), C is a curve from (π, π/2, 2) to (0, π, 1);

(f) F (x, y, z) = (cosx cos (2y) ,−2 sinx sin (2y) , 0), C is a curve from (π/2,−π, 2) to¡π/4, 0,

√3¢;

10. Show thatRC

¡x2y, 2xy2, 0

¢· dr is path dependent in R3.

11. Determine which of the following line integrals are path independent in R3. If affirmative, finda function f such that F = grad f .

(a)RC

¡2xy2, 2x2y, 1

¢· dr

(b)RC (y,−zx, z) · dr

(c)RC (yz, xz, xy) · dr

(d)RC

¡ye2z, 0,−zey

¢· dr

(e)RC

³3 (x+ y)2 , 6 (x+ y)2 , 0

´· dr

(f)RC (e

z, 2y, xez) · dr

12. Evaluate the line integral with respect to arc lengthRC fds in the following cases:

(a) f (x, y, z) = x2 + y2, C : y = 3x, from (0, 0) to (2, 6);

(b) f (x, y, z) = x3y, C : r (t) = (2 cos t, 2 sin t, 0), t ∈ [0, π/2];(c) f (x, y, z) = x2 + y2 + z2, C : r (t) = (cos t, sin t, 2t), t ∈ [0, 4π];(d) f (x, y, z) =

p2 + x2 + 3y2, C : r (t) =

¡t, t, t2

¢, t ∈ [0, 3];

(e) f (x, y, z) = 1 + y2 + z2, C : r (t) = (t, cos t, sin t), t ∈ [0, π];(f) f (x, y, z) = x2 + 3

√xy, C : r (t) =

¡cos3 t, sin3 t, t

¢, t ∈ [0, π].

13. Determine the mass and the coordinates of the center of mass of the following curves (assumemass density is equal to 1):



(a) Arc of the circle of radius R with center the origin lying in the second quadrant (i.e. x ≤ 0and y ≥ 0);

(b) Line segment from (1, 2, 3) to (4, 5, 6);

(c) Part of the parabola y = x2 − 2x− 3 between (−1, 0) and (3, 0).

14. Redo the previous exercise for a variable mass density f (x, y, z) = x.

15. Describe the given region D as the region between the graphs of two functions:

a) D = {(x, y) : 0 ≤ x ≤ 3,−1 ≤ y ≤ 1}b) D =

©(x, y) : y ≥ x2 − 9, y ≤ 2x− 1

ªc) D =

©(x, y) : x2 + y2 ≤ 2x

ªd) D =

n(x, y) : (x− 1)2 + y2 ≤ 4, (x+ 1)2 + y2 ≤ 4

oe) D =

½(x, y) : x2 + y2 ≤ 4, x2 + 1

4y2 ≥ 1, x ≥ 0

¾f) D : ∆ABC, where A (−1, 0) , B (2, 0) , C (0, 3)

16. Write the double integralR R

D f(x, y)dxdy as an iterated integral for the domains D given is inthe previous exercise.

17. Evaluate the given double integrals

a)

ZZD

x sin (xy) dxdy; D = {(x, y) : x ∈ [0, 1] , y ∈ [0, π]}

b)

ZZD

¡x2 + y2dxdy

¢, D : 4ABC, where A (1, 1) , B (2,−1) , C (2, 1)

c)

ZZD

xy2dxdy, D = {(x, y) : 0 ≤ x ≤ 2; 0 ≤ y ≤ ex}

d)

ZZD

xy2dxdy, D =©(x, y) : x2 + y2 ≤ 2x

ªe)

ZZD

xydxdy, D =©y ≥ x2 − 9, y ≤ 2x− 1, x ∈ [−2, 4]

ªf)

ZZD

xydxdy, D =©y ≥ x2 − 9, y ≤ 2x− 1, x ∈ [−2, 3]

ª18. Find the volume of the regions indicated below

a) x2 + y2 + z2 ≤ R2

b) z ≥ x2 + y2, z ≤ 3

c)x2

a2+

y2

b2+

z2

c2≤ 1

19. Evaluate the given double integrals by using a change of variables

a)

ZZD

xdxdy, D =©(x, y) : x2 + y2 ≤ 4, y ≤

√3x, x, y ≥ 0

ªb)

ZZD

ln¡x2 + y2

¢x2 + y2

dxdy, D :©(x, y) : 1 ≤ x2 + y2 ≤ 4

ª.

c)

ZZD

(x+ y) dxdy, D :©(x, y) : x2 + y2 ≤ x+ y

ª.

d)R R

D (x+ y) dxdy, D is the parallelogram with vertices (0, 0) , (4, 0), (1, 3), (5, 3)



3. Calculate the area, the mass and center of mass of the domain D, where the mass density isgiven by f (x, y) = x:

a) D : {(x, y) : x ∈ [−1, 1] , y ∈ [0, 2]}b) D : 4ABC, where A (1, 1) , B (1, 4) , C (4, 1)

c) D :n(x, y) : (x− 1)2 + (y − 1)2 ≤ 1

od) D =

n(x, y) : x

2

a2+ y2

b2≤ 1

o


Integrability 7

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