Andrew Rosen

Chapter 11 - Vectors and Vector-Valued Functions (using LATEX)

Scalar Multiplication:

Given a scalar, c, and a vector, −→v , the scalar multiple c−→v is a vector whose magnitude is |c| multiplied bythe magnitude of −→v

If c < 0, then c−→v and −→v point in opposite directionsTwo vectors are parallel if they are scalar multiples of one another

Because 0−→v = 0 for all vectors, the zero vector is parallel to all vectors

Vector Components:

Round brackets, ( ), are for coordinates and angled brackets, 〈 〉, are for components of a vector.If −→v = 〈a, b, c〉, then |−→v | =

√a2 + b2 + c2

Unit Vectors:

A unit vector is a vector of magnitude (length) 1

If −→v is not the zero vector, then the unit vector in the direction of −→v is de�ned as: v =−→v|−→v |

Also, ±−→v|−→v |

are unit vectors parallel to −→v

Unit Vectors: i = 〈1,0,0〉, j = 〈0,1,0〉, k = 〈0,0,1〉

Vectors in Three Dimensions:

If the curled �ngers of the right hand are rotated from the positive x axis to the positive y axis, the thumbis in the direction of the positive z axis

To �nd−−→PQ from P (2, 1, 2) and Q (4, 3, 6), the values must be subtracted since it's the vector between

these points.

Distance Formula in xyz-Space:√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

A sphere is de�ned as: (x− a)2 + (y − b)2 + (z− c)2 = r2

Example: (Note: Uses �completing the square,� which entails dividing the middle component by 2, squaringit, and adding and subtracting it)

x2 + y2 + z2 + 4x− 6y + 2z = −6(x2 + 4x+ 4− 4) + (y2 − 6y + 9− 9) + (z2 + 2z + 1− 1) = −6(x+ 2)2 − 4 + (y − 3)2 − 9 + (z + 1)2 − 1 = −6(x+ 2)2 + (y − 3)2 + (z + 1)2 = 8

Dot Products:

If −→u = 〈u1, u2, u3〉 and −→v = 〈v1, v2, v3〉, then the dot product is: −→u · −→v = u1v1 + u2v2 + u3v3

It can also be de�ned as −→u · −→v = |−→u ||−→v | cos(θ) if θ is the angle between the 2 vectorsVectors are said to be orthogonal if the dot product of the vectors is zero

Projections:

To de�ne the projection of −→u onto −→v , the following notation is used:−−−→Proj−→v

−→uThe scalar projection is Scal−→v

−→u = |−−−→Proj−→v

−→u |

Scal−→v−→u = |−→u | cos(θ) =

−→u · −→v|−→v |

−−−→Proj−→v

−→u =

(−→u · −→v|−→v |

) −→v|−→v |

To �nd the projection of −→u orthogonal to −→v , this equation would be used: −→u −−−−→Proj−→v

−→u

1

Note: The projection of a vector on another vector is merely a scalar multiple of the vector its being projectedonto. It doesn't even have to be in the same direction (can be antiparallel)

Cross-Product:

|−→u ×−→v | = |−→u ||−→v | sin(θ) where θ is between 0 and π radians (angle between −→u and −→v ) and thedirection of the cross product is indicated by the right hand rule

Note: Cross product produces vector orthogonal to both vectors being crossedThe magnitude of the cross product is also the area of a parallelogram with −→u and −→v as the sidesAnti-Commutative: −→u ×−→v = −(−→v ×−→u )i→ j → k (so, i× j = k, and j × k = i, and k × i = j

−→u ×−→v = (u2v3 − u3v2)i+ (u3v1 − u1v3)j + (u1v2 − u2v1)kTo get this, set up a 3x3 matrix:

−→u ×−→v =

∣∣∣∣∣∣i j ku1 u2 u3v1 v2 v3

∣∣∣∣∣∣ = i

∣∣∣∣ u2 u3v2 v3

∣∣∣∣− j ∣∣∣∣ u1 u3v1 v3

∣∣∣∣+ k

∣∣∣∣ u1 u2v1 v2

∣∣∣∣Then do the simple diagonal rule determinant for the 2x2 matrices (SE arrow multiplication minus SWarrow)

Area of a Parallelogram with sides P,Q,R, S: |−−→PQ×

−→PR|

Area of a Triangle with sides P,Q,R:|−−→PQ×

−→PR|

2

Vector-Valued Functions:

Equation of a Line passing through 〈x0, y0, z0〉 the direction ‖ to −→v = 〈v1, v2, v3〉 is:−→r (t) = 〈x0 + tv1, y0 + tv2, z0 + tv3〉Also can be written parametrically as: x = x0 + tv1, y = y0 + tv2, z = z0 + tv3What is needed is a point on the line and a vector that's parallel

There are in�nite vector-valued functions for a curve/line because the parameter can be changed usinganother pointTo de�ne a line segment, a speci�c domain for the parameter must be written along with

the vector-valued function for the whole line, which can be determined by doing the �nal

coordinate minus the initial coordinate and making it a vector that will become −→v or 〈v1, v2, v3〉To �nd the domain, look for all values of t that will give a value at the beginning and end of the line via

the given coordinatesEquation of a curve: −→r (t) = 〈f(t), g(t), h(t)〉, a< t< b

The domain of t is all t values that make f(t), g(t), h(t) de�nedTo graph a function, make z zero to plot in the xy − plane and then make a new plot adding z in

Finding intersections of planes and curves is straightforward, so keep it that way:Ex: y − x = 0 and −→r (t) = 〈15 cos(t), 15 sin(t), t〉15 sin(t)− 15 cos(t) = 0 and solve for t then plug into the original

Calculus with Vector-Valued Functions:

As long as r′(t) 6= −→0 , then r′(t) is the tangent vector assuming r(t) is fully di�erentiable on the interval

The Unit Tangent Vector:−→T =

r′(t)

|r′(t)|All derivative rules still apply, but be careful:

Dot Product Rule:d

dt(−→u · −→v ) = −→u ′ · −→v +−→u · −→v ′

Cross Product Rule:d

dt(−→u ×−→v ) = −→u ′ ×−→v +−→u ×−→v ′

Integration is as usual, but the constant is a vector (−→C ) and can be added to the whole function or its

components. Also, integration can be distributed in to the components:´ −→r (t)dt = −→R (t) +−→C

2

Motion in Space:−→r ′′(t) = −→v ′(t) = −→a (t)

When integrating and �nding the constants using initial values (t = 0) problems, remember that theconstant C is in more than one dimension.Speed = |−→v (t)|

Arc Length:

S =´ ba

√f ′(t)2 + g′(t)2 + h′(t)2dt =

´ ba|−→r ′(t)|dt

If a vector is multiplied by a scalar multiple, it is the same equation even though it is di�erent parametrization.Even so, its arc length is identical.Circular Motion with constant |−→r |: −→r · −→v = 0

Reparameterization:

If a curve or function has the dummy variable t replaced with one that is respect to arc length, it is uniqueand has meaning (distance on curve)

S(t) =´ ta|−→r ′(u)|du

Find S(t) and then solve for t to substitute into the vector-valued functiona is simply the lowest point on the domain

Change the domain for S(t) by replacing t for the function in terms of s in the inequality for the domainand then isolate s

If |−→r ′(t)| = 1, then arc length was already used for parameterization

3

Andrew Rosen

Chapter 12 - Functions of Several Variables (LATEX)

Planes and Surfaces1:

The equation for a plane that contains (x0, y0, z0) and has a normal vector −→n = 〈a, b, c〉 is:a(x− x0) + b(y − y0) + c(z − z0) = 0

This equation is unique because it is set to zero

For an equation, the point on the plane and a vector normal to the plane are needed

It can be arranged to make ax+ by + cz = d and the coe�cients of the variables are −→nNote: Every plane has an orientation determined by −→n

Finding Equation of a Plane:

1) Create two vector segments via subtraction

2) Cross the two new vectors since the cross product is perpendicular

3) Plug into equation

3) Isolate constants and variables

Example: A plane through three points

Find the equation of a plane that contains P (2,−1, 3), Q(1, 4, 0), and R(0,−1, 5):Step 1:

−−→PQ = 〈−1, 5,−3〉 and

−→PR = 〈−2, 0, 2〉

Step 2:−−→PQ×

−→PR =

i j k−1 5 −3−2 0 2

= 〈10, 8, 10〉

Step 3: 10(x− 2) + 8(y − (−1)) + 10(z − 3) = 0

Step 4: 5x+ 4y + 5z = 21

Angle of Intersecting Planes:

1) The two normal vectors of each plane produce an angle identical to the angle of intersection

2) Solve for cos(θ) between the two normal vectors using the dot product equation

Planes are parallel if their respective normal vectors are parallel (scalar multiples)

Planes are orthogonal if their respective normal vectors are orthogonal (dot product is zero)

Example: Intersecting Planes

Find the equation of the line of intersection of the planes Q : x+ 2y + z = 5 and R : 2x+ y − z = 7

Step 1: Set z = 0 and solve for x and y (x = 3, y = 1, z = 0)

Step 2: Cross the two normal vectors (−→v = −→n 1 ×−→n 2)

Step 3: Write an equation of a line from these two steps

Distance between a Point and a Plane:

If given a point P and an equation for a plane, do the scalar projection of−−→PQ (where Q is a point on the

plane) projected onto the normal vector of the plane, −→n . Therefore, this is scal−→n−−→PQ.

1It is best to be familiar with how to to do all examples of planes in Chapter 12.1

1

Distance between a Point and a Line:

If given a point P and the equation for a line, set t = 0 to �nd another point, Q. Then �nd the parallel

vector from the equation of the line, −→v . Then cross−−→PQ and −→v . Finally, divide this by |−→v |

Cylinders:

Think of a cylinder as an extruded spline (with surfsculpt)

If an equation is �missing� a variable, it is a cylinder (eg: z = x2 in R3)

Not all cylinders have this property though

Cylinders extend in the dimension that is �missing� for in�nity and negative in�nity

Graphs and Level Curves:

Domain and range can be de�ned as usual

Ex: f(x, y) =√9− x2 − y2

D = {(x, y) ∈ R2|x2 + y2 ≤ 9}R = [0, 3]

Level Curve: f(x, y) = c where c is a constant

Ex: f(x, y) =√9− x2 − y2 = c

x2 + y2 = 9− c2 ∴ c is between 0 and 3 (range)

Partial Derivatives:

A partial derivative (∂) simply is the computation of a derivative that is stated while the other variables aremade as constants

Ex: z = 2x sin(y)∂z

∂x= 2 sin(y)

Higher Order:∂2f

∂x2= fxx =

∂

∂x(fx)

A partial derivative is di�erentiable if all of its (�rst order) partial derivatives are continuous at apoint (a, b)

fxy = fyx for continuous functions

The Chain Rule:

To do the chain rule, a tree diagram should be set up with all of the variables. From this, one can trace outthe paths to get to the desired variable. Multiply partial derivatives down a chain, and add chains together.

Implicit Di�erentiation:

dy

dx= −

Fx

Fy

To implicitly di�erentiate, compute partial derivatives but be careful of implicitly de�ned functions

2

Eg:∂z

∂xof x3 + y3 + z3 + 6xyz = 1 where z is implicit for x and y

3x2 + 3z2∂z

∂x+ (6yz + 6xy

∂z

∂x) = 0 ← z is implicitly de�ned for x and is not constant

∂z

∂x=−3x2 − 6yz

3z2 + 6xy

Directional Derivative:−→∇f = 〈fx, fy, fz〉

D−→u f =−→∇f · −→u (where −→u is a unit vector 〈a, b, c〉 from

−→u|−→u |

)

Since−→∇f · −→u = |

−→∇f ||−→u | cos(θ), θ = 0◦ is the largest value because |−→u |=1

Maximum ascent: u =

−→∇f

|−→∇f |

∴ u is in the direction of maximum derivative (always in direction of the

gradient)

Value of Maximum Ascent: |−→∇f |

To �nd maximum descent, negate the x and y components of the unit vector−→∇ is always ⊥ to level curves

Tangent Planes and Linear Approximations:

Tangent Plane for F (x, y, z) = 0: Fx(x0, y0, z0)(x− x0) + Fy(x0, y0, z0)(y − y0) + Fz(x0, y0, z0)(z − z0) = 0,where F is di�erentiable at the point P0(x0, y0, z0) and z = F (x, y)

Tangent Plane for z = f(x, y) : z = Fx(x0, y0)(x− x0) + Fy(x0, y0)(y − y0) + F (x0, y0)

Linear Approximation: L(x, y) = Fx(x0, y0)(x− x0) + Fy(x0, y0)(y − y0) + F (x0, y0)

Di�erentials: dz = Fx(x, y)dx + Fy(x, y)dy

Maximum/Minimum Problems:

Critical Point:

1) fx(a, b) = 0 and fy(a, b) = 0

Set both partial derivatives equal to zero simultaneously and solve (don't forget ± with even roots)

2)Fx or FyDNE

Discriminant: D(a, b) = fxx(a, b)fyy(a, b)− (fxy(a, b))2

1. If D > 0 and fxx(a, b) > 0, then f(a, b) is a local minimum

2. If D > 0 and fxx(a, b) < 0, then f(a, b) is a local maximum

3. If D < 0 then f(a, b) is a saddle point

4. If D = 0 it's inconclusive

Closed Set: Set of points that contains all its boundary points

Bounded Set: Set of points that is contained within some bigger �nite circle

Extreme Value Theorem: If f is continuous on a closed, bounded domain, D, then f attains an absolutemax and absolute min value for some point(s) in D

3

Absolute Extrema:

1. Find the critical points of f in D (no need to do second derivative test) and see if it's inside boundary

2. Find extreme value on the boundary

3. The largest value of f from steps 1 and 2 is the absolute max and vice versa

Note: If the closed region is that of a circle (eg: x2 + y2 ≤ 25), then it can be modeled with x = a cos(θ)and y = a sin(θ) where a is the disk's radius

Note: To �nd the shortest distance between two regions, you can use the distance formula. Doing partialderivatives on it is nasty, so square both sides and then �nd the critical points as usual. Then plug the xand y values into the original function (not the squared one)

Lagrange Multipliers:

Find all (x, y, z) and λ such that−→∇f(x, y, z) = λ

−→∇g(x, y, z)

If this is a physical problem, λ usually cannot equal zero and thus can be divided (same with x, y, z)

Note: If the boundary is given as a function with an equals sign, do not compute the critical point insidethe region. Only compute it on the boundary. If the boundary is given as an inequality, compute the criticalpoint inside the region as well as the points on the boundary.

4

Quadric Surfaces:

5

Andrew Rosen

Chapter 13 - Multiple Integration (LATEX)

13.1 - Double Integration over Rectangular Regions

Fubini's Theorem: If f is continuous on a rectangle with [a, b]× [c, d] = D, then

¨

D

f(x, y) dA =

ˆ ba

ˆ dc

f(x, y) dy dx =

ˆ dc

ˆ ba

f(x, y) dx dy

Example: (Note - if it's´−2x− y + 6 dy, then it becomes −2xy − y2

2 + 6y + C)

´ 20

´ −1−2 (9x+ y) dx dy →

´ 20

(92x

2 + xy∣∣−1−2

)dy →

´ 20

(−272 + y

)dy → −27

2 y + y2

2

∣∣∣20→ −25

The average value of an integrable function f over a region D is:

f =

˜D

f(x, y) dA

Area of D=

˜D

f(x, y) dA

˜D

dA

Remember: If you are looking for the bounds on an integral with respect to dx, the lower bound is the lowerfunction and the upper bound is the higher function. If you are looking for the bounds on an integral withrespect to dy, the lower bound is the left-most function and the upper bound is the right-most function,assuming that this is on a typical xy plane.

13.2 - Double Integrals over General Regions

With integrals over nonrectangular regions, the order of integration cannot be simply switched. A correctstatement would be the following where the bounds are x1 = a, x2 = b, y1 = c, and y2 = d:

¨

D

f(x, y) dA =

ˆ ba

ˆ h(x)g(x)

f(x, y) dy dx

¨

D

f(x, y) dA =

ˆ dc

ˆ h(y)g(y)

f(x, y) dx dy

How to �nd the bounds of a double integral1:

1. Determine with which variable the inner integral is with respect to: dx or dy

2. The bounding curves determine the limits of integration for the variable determined in the �rst step

3. The bounds of the remaining variable is the projection of the region on that axis

1It is important to be able to identify the standard quadric surfaces

1

Example: Double integral with dx on the inside and dy on the outside

Compute the iterated integral of the region bounded by y = 14− x, y = 3, and x = 4 with dx on the insideand dy on the outside

Step 1: Plot the region

Step 2: Determine the bounds for the dx integral

4 ≤ x ≤ −y + 14

Step 3: Determine the bounds for the dy integral

3 ≤ y ≤ 10

Step 4: Write the iterated integral:

ˆ 10

3

ˆ −y+14

4

f(x, y) dx dy

Example: Double integral with dy on the inside and dx on the outside

Compute the iterated integral of the region bounded by y = 14− x, y = 3, and x = 4 with dy on the insideand dx on the outside

Step 1: Plot the region (see previous plot)

Step 2: Determine the bounds for the dy integral

3 ≤ y ≤ 14− xStep 3: Determine the bounds for the dx integral

4 ≤ x ≤ 11

Step 4: Write the iterated integral:

ˆ 11

4

ˆ 14−x

3

f(x, y) dy dx

The volume between two surfaces where g(x, y) ≥ f(x, y) is:

V =

¨

D

(g(x, y) − f(x, y)) dA

If D is a region in the xy plane then the area of that region is:

AreaD =

¨

D

dA

2

13.3 - Double Integrals in Polar Coordinates2

Before doing double integration in polar coordinates, it is essential to recall the following identities:

r =√x2 + y2 → r2 = x2 + y2

x = r cos(θ), y = r sin(θ)

As long as β − α ≤ 2π while α ≤ θ ≤ β and 0 ≤ a ≤ r ≤ b, then the following is the double integral over apolar rectangular region for a function, f(x, y):

¨

D

f(r, θ) dA =

ˆ θ=βθ=α

ˆ r=br=a

f [r cos(θ), r sin(θ)] r dr dθ

With double integrals over non-rectangular polar regions, the order of integration cannot be switched directly.A more general expression for the double integral over a general polar rectangular region where 0 ≤ g(θ) ≤r ≤ h(θ) and β − α ≤ 2π with α ≤ θ ≤ β for a function, f(x, y), is:

¨

D

f(r, θ) dA =

ˆ θ=βθ=α

ˆ r=h(θ)r=g(θ)

f [r cos(θ), r sin(θ)] r dr dθ

Note: Don't forget to multiply the integrand by a factor of r!

Area of Polar Regions:

A =

¨

D

dA =

ˆ θ=βθ=a

ˆ r=h(θ)r=g(θ)

r dr dθ

The following trigonometric identities are crucial to integration of squared trigonometric functions:

sin2(θ) =1

2(1− cos(2θ))

cos2(θ) =1

2(1 + cos(2θ))

2It is essential to be able to recall the typical trigonometric values for sine, cosine, and tangent

3

Example:

Set up an equation to �nd the area of 1 leaf of the rose, r = cos(2θ)

1) Plot the function:

2) Write the bounds

0 ≤ r ≤ cos(2θ) and −π4 ≤ θ ≤π4

Note: When writing the bounds of integration for θ, make sure that it is going from a lower value to a highervalue and make sure it is the correct region of the boundary. In this case, π4 to 7π

4 would be incorrect sinceit'd sweep more than one leaf.

3) Set up double integral

A =

ˆ π4

−π4

ˆ cos(2θ)

0

r dr dθ

13.4 - Triple Integrals

˚

D

f(x, y, z) dV =

ˆ ba

ˆ h(x)g(x)

ˆ H(x,y)

G(x,y)

f(x, y, z) dz dy dx

Note: Five other orders of integration could be set up based on Fubini's Theorem

How to Find the Bounds of Integration:

1. Imagine that you really, really hate calculus, and triple integrals make you want to stab things. Thisattitude will help you greatly!3

2. Your �rst set of limits can be �gured out if you imagine (fatally) �stabbing� the three-dimensionalboundary in the direction of the axis of the variable you're integrating. Find where the knife (or spear,rusty lance, extra sharp pencil, etc.) enters the three-dimensional boundary for the �rst time andwhere it exits. Basically, �nd the entrance and exit wounds. Write this as an inequality (eg: a ≤ z ≤ b,where the integral is with respect to dz).

3. Set the �rst variable of integration to zero (so, if you integrated with respect to z �rst, make z = 0) tocreate a plane. It might even be helpful to draw a new graph in a normal 2-variable Cartesian plane.Now stab in the direction of the middle variable of integration. Find the entrance and exit wounds.

4. Now you are left with one variable to �nd the limits of integration for. The bounds of the remainingvariable is the projection of the region on that axis.

3Don't worry. I'm not really that violent.

4

How to switch the bounds of a triple integral (and how to graph a 3D function given the bounds):

1. Since the middle and outer integrals' bounds are the projection of the 3D region, plot this 2D projection�rst. Be careful of this major fact: If you have

˝D

dx dy dz, you'd graph a yz projection �rst. If one

of the bounds for, let's say, y is something like y = z then you can't graph that directly! See what zequals and substitute that in for y = z.

2. From here, the bounds for the middle integral can be found as usual. The bounds for the outer integralare the projection of the 2D graph on the axis of the variable that the outer integral is with respect to(make sure that the outer integral has constants for the bounds)

3. Now, extrude this projection in the dimension of the inner integral variable. The bounds for the innerintegral can then be found as normal.

4. The key is to realize which bounds are functions of which variables and to adjust them accordingly.

5

Example:

Set up a triple integral to �nd the volume of the given solid region in the �rst octant bounded by the plane12x+ 16y + 12z = 48 and the coordinate planes.

Given graph of the boundary:

Step 1: Pick an order of integration. I will pick dz dy dx arbitrarily. Any order can be used.

Step 2: Imagine stabbing this boundary in the z axis. It enters at z = 0 and exits at the equation of the

plane solved for z, which is 4− x− 4y

3. Therefore, 0 ≤ z ≤ 4− x− 4y

3Step 3: Set z = 0 to create a new Cartesian plane. Re-plotting the boundary might be helpful. The graphis shown below.

Step 4: Find the boundaries for y. The hypothetical knife would enter at y = 0. Now you need the exitpoint. There are two easy ways to do this. One is to simply solve for y when z = 0. This would make

y = −3

4x + 3. Another way to do this, which is equally as easy, is to recognize that the slope of a line is

the∆y

∆x, which is −3

4. This, in conjunction with the knowledge that the y intercept is (0, 3) can yield the

equation of the line, which is −3

4x+ 3. Therefore, 0 ≤ y ≤ −3

4x+ 3.

Step 5: Find the remaining boundary for x. The projection of the previous boundary on the x axis yields0 ≤ x ≤ 4.

Step 6: Use this information to piece together a triple integral. It would thus be:

ˆ 4

0

ˆ − 3x4 +3

0

ˆ 4−x− 4y3

0

dx dy dz

6

13.5, Part I - Triple Integrals in Cylindrical Coordinates

The triple integral of f over D in cylindrical coordinates is:

˚

D

f(x, y, z) dV =

ˆ βα

ˆ h(θ)g(θ)

ˆ H(r cos(θ),r sin(θ))

G(r cos(θ),r sin(θ))

f(r, θ, z) dz r dr dθ

The following are important conversion rules between rectangular and cylindrical coordinates:

1. x = r cos(θ)

2. y = r sin(θ)

3. z = z

Example:

Set up, but do not evaluate,´ 20

´√4−x2

−√4−x2

´ 2√x2+y2

(x2 + y2) dz dy dx in cylindrical coordinates.

Step 1: Plot the boundary region. The tip of the cone is at (0, 0, 0) and opens from z = 0 to z = 2

Step 2: Find the boundaries of z by converting to cylindrical coordinates. The boundaries become r ≤ z ≤ 2.Also note that the integrand itself becomes r2.

Step 3: Collapse the cone so that z = 0 and plot it (indicated by shaded region)

Remember to check the boundaries given

Step 4: Find the boundaries for r and θ. With this graph, it is clear that 0 ≤ r ≤ 2. The angle must sweepfrom a smaller angle to a higher angle for the boundary, so −π2 ≤ θ ≤

π2 .

Step 5: Set up the triple integral. It would become, with the boundaries already determined:

ˆ π2

−π2

ˆ 2

0

ˆ 2

r

r2 r dz dr dθ

13.5, Part II - Triple Integrals in Spherical Coordinates

There are three new coordinates to recognize in spherical coordinates:

1. ρ is the distance from the origin to a point, P

2. φ is the angle between the positive z-axis and an arbitrary line OP that goes from 0 to π4

3. θ is the same angle as in cylindrical coordinates and measures rotation about the z-axis relative to thepositive x-axis

4φ = 0 is a line from the origin going upward vertically and φ = π is a line from the origin going downward vertically

7

The triple integral of f over a region, D, in spherical coordinates is:

˚

D

f(ρ, φ, θ) dV =

ˆ βα

ˆ ba

ˆ h(φ,θ)g(φ,θ)

f(ρ, φ, θ)ρ2 sinφdρ dφ dθ

The following are important conversion rules between rectangular and spherical coordinates:

1. x = ρ sinφ cos θ

2. y = ρ sinφ sin θ

3. z = ρ cosφ

Also note: ρ =√r2 + z2 and ρ2 = x2 + y2 + z2 = r2 + z2

Example:

Set up, but do not evaluate, the following integral in spherical coordinates:˝D

e−(4x2+4y2+4z2)

32 dV where

D is a sphere of radius 6.

Step 1: Convert known functions to spherical coordinates. The triple integral becomes˝D

(e−4r

2) 3

2

dV

Step 2: Find the bounds for ρ. This will extend from 0 to the sphere, which is 6

Step 3: Find the bounds for φ. The bounds for this will be from 0 to π. The reason that it is not from 0 to2π is because the bounds of θ take care of that part of the sweeping action

Step 4: Find the bounds for θ. The bounds for this will be one full revolution from 0 to 2π

Step 5: Set up the triple integral. It will be:

ˆ 2π

0

ˆ π

0

ˆ 6

0

e−8ρ3

ρ2 sinφdρ dφ dθ

Example:

Set up, but do not evaluate, an expression for the volume of the smaller region cut from the solid sphereρ ≤ 14 by the plane z = 7.

Step 1: Find the bounds for ρ. An equation for a plane is given by a secφ because z = 7 is the plane andz = ρ cosφ, so the lower bound for ρ is 6 secφ, and the upper bound is 14.

Step 2: Find the bounds for φ. The lower limit is φ = 0, and the upper limit ofπ

3can be found by setting

7

cosφequal to 14 and solving for φ.

Step 3: Find the bounds for θ. These are simply those of a circle in order to create one full revolution.Therefore, the lower bound is 0 and upper bound is 2π.

Step 4: Write out the triple integral.

ˆ 2π

0

ˆ π/3

0

ˆ 14

7/ cosφ

ρ2 sinφdρ dφ dθ

8

9

Andrew Rosen

Chapter 14 - Vector Calculus (LATEX)

14.1 - Vector Fields

A vector �eld is written as−→F (x, y, z) = 〈f(x, y, z), g(x, y, z), h(x, y, z)〉 because for every point in space

(x, y, z) you get a di�erent vector

A vector �eld is conservative if it is the gradient (−→∇) of a scalar function

There exists a �potential function�, φ(x, y, z), such that−→F (x, y, z) =

−→∇φ(x, y, z)

−→F would be considered a gradient �eld

14.2 - Line Integrals

If f is a continuous function on the smooth curve C, −→r (t) = 〈x(t), y(t), z(t)〉, and a ≤ t ≤ b, then thefollowing is a line integral1:

ˆ

C

f(x, y, z) ds =

ˆ b

a

f(x(t), y(t), z(t)) |−→r ′(t)| dt =ˆ b

a

f(x(t), y(t), z(t))

√[x′(t)]

2+ [y′(t)]

2+ [z′(t)]

2dt

How to Evaluate a Line Integral:

1. Find a parametric description of C in the form −→r (t) = 〈x(t), y(t), z(t)〉 for a ≤ t ≤ b

2. Compute |−→r ′(t)| =√

[x′(t)]2+ [y′(t)]

2+ [z′(t)]

2

3. Make substitutions for x and y in the integrand and evaluate an ordinary integral:´C

f(x, y, z) ds =

´ baf(x(t), y(t), z(t)) |−→r ′(t)| dt

To �nd the length of a curve, −→r (t), on the bounds a ≤ t ≤ b, you can do the following line integral:´ ba|−→r ′(t)| dt

Example:

Set up the following line integral in an �ordinary� integral :´C

(2 + x2y) ds, where C is the upper half of the

unit circle

Step 1: Write a parametric description of the curve. A circle can be written in terms of cosine and sine, soit is −→r (t) = 〈cos t, sin t〉, where 0 ≤ t ≤ πStep 2: Substitute in for x and y. The function becomes f(x(t), y(t), z(t)) = 2 + cos2 t sin t

Step 3: Compute the magnitude of the derivative of−→r (t). −→r ′(t) = 〈− sin t, cos t〉, so |−→r ′(t)| =√

sin2 t+ cos2 t =1

Step 4: Substitute |−→r ′(t)| in for ds. This becomes: ds = |−→r ′(t)| dt = 1 dt = dt

Step 5: Put all the pieces together to evaluate an ordinary integral:

ˆ

C

(2 + x2y) ds =

ˆ π

0

(2 + cos2 t+ sin t) dt

1If C is not smooth everywhere (piece-wise), the line integral can still be evaluated if the curve is broken up into segments

1

Line Integrals of Vector Fields:

If−→F (x, y, z) is a continuous vector �eld de�ned on a piece-wise smooth curve, −→r (t), from a ≤ t ≤ b, the line

integral of−→F on C is:

ˆ

C

−→F · d−→r =

ˆ

C

(−→F · T ) ds =

ˆ b

a

−→F (−→r (t)) · r′(t) dt

Let−→F be a continuous vector �eld and let C be a closed smooth curve. The circulation is the same as the

previously de�ned line integral of a vector �eld:

Circulation =

˛

C

F =

ˆ

C

−→F · T ds =

ˆ b

a

−→F (−→r (t)) · r′(t) dt

Let−→F be a continuous vector �eld and let C be a closed smooth curve with counterclockwise orientation.

The (outward) �ux is:

F lux =

ˆ

C

−→F · n ds =

ˆ b

a

(f y′(t)− g x′(t)) dt

Note: n = T × kNote: A positive answer means a positive outward �ux

14.3 - Conservative Vector Fields

A vector �eld is said to be conservative on a region if there exists a scalar function, φ, such that−→F =

−→∇φ

For a conservative vector �eld,−→F ,¸C

−→F · d−→r = 0 on all simple closed smooth oriented curves C

In R2,∂f

∂y=∂g

∂xfor conservative vector �elds

In R3,∂f

∂y=∂g

∂x,∂f

∂z=∂h

∂x, and

∂g

∂z=∂h

∂yfor conservative vector �elds

How to �nd φ in 2-space:

1. Integrate φx = f with respect to x to obtain φ, which includes an arbitrary function c(y)

2. Compute φy via partial di�erentiation and equate it to g to obtain an expression for c′(y)

3. Substitute in for g and integrate with respect to y to �nd c(y)

4. Put it all together for an expression of φ

How to �nd φ in 3-space:

1. Integrate φx = f with respect to x to obtain φ, which includes an arbitrary function c(y, z)

2. Compute φy via partial di�erentation and equate it to g to obtain an expression for cy(y, z)

3. Substitute in for g and integrate cy(y, z) with respect to y to obtain c(y, z), including an arbitraryfunction d(z)

4. Compute φz via partial di�erentiation (using φ from step 1 and d(z) substituted in from step 3) andequate it to h

5. Substitute in for h and integrate with respect to z to get d(z)

2

Example:

Determine whether the following vector �eld is conservative on R2. If so, determine the potential function

when−→F = 〈3x, 3y〉

Step 1: Prove if−→F is conservative

f(x, y) = 3x and g(x, y) = 3y

∂f

∂y= 0 and

∂g

∂x= 0, therefore it is conservative

Step 2: Find φ

φx = f = 3x,´φx dx =

´3x dx =

3

2x2 + c(y)

∂

∂y

(3

2x2 + c(y)

)= c′(y) = g

c′(y) = 3y´c′(y) dy =

´3y dy =

3

2y2 + C

Step 3: Put it together

φ(x, y) =3

2x2 +

3

2y2 + C

Example:

Determine whether the following vector �eld is conservative on R3. If so, determine the potential function

when−→F = 〈6y + z, 6x+ 7z, x+ 7y〉

Step 1: Prove if−→F is conservative

f(x, y, z) = 6y + z and g(x, y, z) = 6x+ 7z and h(x, y, z) = x+ 7y

∂f

∂y= 6 =

∂g

∂x,∂f

∂z= 1 =

∂h

∂x, and

∂g

∂z= 7 =

∂h

∂y, therefore it is conservative

Step 2: Find φ

φx = f = 6yz

φ =´φx dx =

´6y + z dx = 6xy + xz + c(y, z)

φy =∂

∂y(6xy + xz + c(y, z)) = 6x+ cy(y, z) = g

6x+ cy(y, z) = 6x+ 7z → cy(y, z) = 7z

c(y, z) =´cy(y, z) dy =

´7z dy = 7yz + d(z)

φz =∂

∂z(6xy + xz + 7yz + d(z)) = x+ 7y + d′(z) = h

x+ 7y = x+ 7y + d′(z)→ d′(z) = 0

d(z) =´d′(z) dz =

´0 dz = C

Step 3: Put it together

φ(x, y, z) = 6xy + xz + 7yz + C

3

In order to evaluate the line integral´C

−→∇φ ·d−→r given φ(x, y) and −→r (t), you must take the derivative of −→r (t)

and dot it with−→F (or

−→∇φ since they are equivalent), so make sure the gradient of φ is taken. Then a simple

integral can be calculated.

Fundamental Theorem for Line Integrals:

ˆ

C

−→F · T ds =

ˆ

C

−→F (−→r (t)) · −→r ′(t) dt = φ(B)− φ(A)

Example:

If given −→r (t) = 〈cos(t), sin(t)〉 for π2≤ t ≤ π and φ(x, y) = xy, use the fundamental theorem for line integrals

to evaluate´C

−→∇φ · d−→r

Step 1: Plug each bound into −→r (t) to �nd coordinates A and B

A : 〈cos(π2

), sin

(π2

)〉 = 〈0, 1〉 ∴ x = 0, y = 1 ∴ A = (0, 1)

B : 〈cos(π), sin (π)〉 = 〈−1, 0〉 ∴ x = −1, y = 0 ∴ B = (−1, 0)

Step 2: Use the fundamental theorem of line integrals

φ(x, y) = xy →´C

−→∇φ · d−→r = φ(B)− φ(A) = (−1)(0)− (0)(1) = 0

4

Andrew Rosen

Chapter 14 - Vector Calculus, Part II (LATEX)

14.4 - Green's Theorem

Green's Theorem - Circulation Form:

˛ −→F · d−→r =

˛f dx + g dy =

¨

R

(∂g

∂x−

∂f

∂y

)dA

Note: The partial derivatives that are subtracted in the double integrand are collectively known as thetwo-dimensional curl

Note: From here on out, the positive direction is counterclockwise. If the curve is oriented clockwise, negatethe answer

If the two-dimensional curl of a vector �eld,−→F = 〈f, g〉, is zero then it is said to be irrotational

If the 2-D curl is zero,−→F is conservative (

−→F =

−→∇φ) because

¸ −→F · d−→r will also be zero

Area of a Plane Region by Line Integrals:

˛

C

x dy = −˛

C

y dx =1

2

˛

C

(x dy − y dx)

Green's Theorem - Flux Form:

˛

C

−→F · n ds =

˛f dy − g dx =

¨

R

(∂f

∂x+

∂g

∂y

)dA

Note: The partial derivatives that are added in the double integrand are collectively known as the two-

dimensional divergence

If the two-dimensional divergence of a vector �eld,−→F = 〈f, g〉, is zero then it is said to be source-free

14.5 - Divergence and Curl

div−→F =

−→∇ ·−→F =

∂f

∂x+

∂g

∂y+

∂h

∂z

curl−→F =

−→∇ ×

−→F = 〈

∂h

∂y−

∂g

∂z,∂f

∂z−

∂h

∂x,∂g

∂x−

∂f

∂y〉

Properties of a Conservative Vector Field:

1. There exists a potential function, φ, such that−→F =

−→∇φ

2.´C

−→F · d−→r = φ(B)− φ(A) for all points on a smooth, oriented curve, C, from A to B

3.¸C

−→F · d−→r = 0 on all simple, smooth, close oriented curves C

4. 1−→∇ ×

−→F =

−→0

1Note: Rule #4 only applies if−→F has continuous partial derivatives on all of R3, so if there is a singularity anywhere then

the theorem cannot be used

1

Curl of a Gradient: If−→F is conservative then

−→∇ × (

−→∇φ) = −→0

Divergence of the Curl:−→∇ · (

−→∇ ×

−→F ) = 0

14.6 - Surface Integrals

A general description for a parametric surface is the following,

−→r (u, v) = 〈x(u, v), y(u, v), z(u, v)〉

A general description for parameterizing a cylinder with its axis along the z axis with radius, r:

−→r (θ, z) = 〈r cos θ, r sin θ, z〉

A general description for parameterizing a cone with its vertex at the origin with radius, r, and height, h:

−→r (θ, z) = 〈rzh

cos θ,rz

hsin θ, z〉

A general description for parameterizing a sphere with radius, ρ, centered at the origin2:

−→r (φ, θ) = 〈ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ〉

Surface Integral of Scalar Functions:

¨

S

f(x, y, z) dS =

¨

R

f(x(u, v), y(u, v), z(u, v)) |−→t u ×−→t v| dA

Surface Integral of Vector Fields:

F lux =

¨

S

−→F · n dS =

¨

R

−→F · (−→t u ×

−→t v) dA

Average value of a function on a surface:

Average =

˜R

f(x(u, v), y(u, v), z(u, v)) |−→t u ×−→t v| dA

˜R

|−→t u ×−→t v| dA

2Finding the magnitude of the cross product of the two partial derivative vectors can be di�cult here, but it will usually be

ρ2 sinφ, which is exactly what is used for a spherical coordinate triple integral! This will be dA here. Even though the sphere

is parameterized with spherical coordinates, do not multiply in another factor of ρ2 sinφ. It's a regular, rectangular double

integral with bounds for φ and θ

2

Example for �nding the equation of a tangent plane at a point on a surface:

Find the equation of the tangent plane at (1, 1, 3) of−→t (u, v) = 〈u2, v2, u+ 2v〉

Step 1: Calculate the cross product of the partial derivatives−→t u = 〈2u, 0, 1〉 and −→t v = 〈0, 2v, 2〉−→t u ×

−→t v = 〈2u, 0, 1〉 × 〈0, 2v, 2〉 = 〈−2v,−4u, 4uv〉

Step 2: Solve for the normal vector by �nding u, v and substituting into the cross product

x = 1 = u2 → u = ±1y = 1 = v2 → v = ±1z = 3 = u+ 2v ∴ u, v = 1−→n (1, 1, 3) = 〈−2,−4, 4〉

Step 3: Set up the tangent plane equation at that point

−2(x− 1)− 4(y − 1) + 4(z − 3) = 0→ −2x− 4y + 4z = 6

14.7 - Stokes' Theorem

Assuming that−→F is a vector �eld show components have continuous �rst partial derivatives on S,

˛ −→F · d−→r =

¨

S

(−→∇ ×

−→F ) · n dS

Note: The normal vector for an explicitly de�ned surface z = g(x, y) is 〈−zx,−zy, 1〉 [Might help to solvesome problems]

Note: If the surface is in the xy plane, the unit normal vector is k

14.8 - Divergence Theorem

¨

S

−→F · n dS =

˚

D

−→∇ ·−→F dV

3

Multivariable Calculus Study Guide:

A LATEX Version

Tyler SilberUniversity of Connecticut

December 11, 2011

1 Disclaimer

It is not guaranteed that I have every single bit of necessary information forthe course. This happened to be some of what I needed to know this specificsemester in my course. For example, Stokes’ Theorem is not even mentioned.

2 Vectors Between Two Points

Given : P (x1, y1) & Q(x2, y2)

−−→PQ =

(x2 − x1

y2 − y1

)

3 Vectors in the Plane

let v =

(v1

v2

)& u =

(u1

u2

)

0 =

(00

)

3.1 Simple Operations

cv =

(cv1

cv2

)|v| =

√v2

1 + v22

v + u =

(v1 + u1

v2 + u2

)

1

3.2 Unit Vectors

i =

(10

)& j =

(01

)v = v1i + v2j

3.3 Vectors of a Specified Length∣∣∣∣ cv|v|∣∣∣∣ = |c|

± cv|v|‖ v

4 Vectors in Three Dimensions

4.1 Notes

Everything in the above section can be expanded to three dimensions. Simplyadd another component.

k =

001

4.2 Random Equations

xy-plane {(x, y, z) : z = 0}

xz-plane {(x, y, z) : y = 0}

yz-plane {(x, y, z) : x = 0}

Sphere: (x− a)2 + (y − b)2 + (z − c)2 = r2

5 Dot Product

5.1 Definitions

u · v = u1v1 + u2v2 + u3v3 = |u||v| cos θ

u ⊥ v⇔ u · v = 0

u ‖ v⇔ u · v = ±|u||v|

2

5.2 Projections

The orthogonal projection of u onto v is denoted projvu and the scalar compo-nent of u in the direction of v is denoted scalvu.

projvu = |u| cos θ

(v

|v|

)=(u · v

v · v

)v

scalvu = |u| cos θ =u · v|v|

6 Cross Product

|u× v| = |u||v| sin θ (1)

u ‖ v⇔ u× v = 0

u× v =

u2v3 − u3v2

u3v1 − u1v3

u1v2 − u2v1

Note: u × v is orthogonal to both u and v and the direction is defined by theright-hand rule.

7 Lines and Curves in Space

7.1 Vector-Valued Functions

r(t) = 〈x(t), y(t), z(t)〉

7.2 Lines

〈x, y, z〉 = 〈x0, y0, z0〉+ t〈a, b, c〉, for −∞ < t <∞

7.3 Line Segments

Given : P1(x1, y1, z1) & P2(x2, y2, z2)

−−−→P1P2 = 〈x1, y1, z1〉+ t〈x2 − x1, y2 − y1, z2 − z1〉, for 0 ≤ t ≤ 1

7.4 Curves in Space

r(t) = 〈f(t), g(t), h(t)〉

Equation 1 is also equal to the area of the parallelogram created by the two vectors.

3

7.5 Limits

limt→a

r(t) =⟨

limt→a

f(t), limt→a

g(t), limt→a

h(t)⟩

8 Calculus of Vector-Valued Functions

8.1 Derivative and Tangent Vector

r′(t) = f ′(t)i + g′(t)j + h′(t)k

Note: r′(t) is the tangent vector to r(t) at the point (f(t), g(t), h(t)).

8.2 Indefinite Integral∫r(t) dt = R(t) + C

Note: C is an arbitrary constant vector and R = F i +Gj +Hk.

8.3 Definite Integral∫ b

a

r(t) dt =

[∫ b

a

f(t) dt

]i +

[∫ b

a

g(t) dt

]j +

[∫ b

a

h(t) dt

]k

9 Motion in Space

9.1 Definitions

a(t) = v′(t) = r′′(t)

Speed = |v(t)|

9.2 Two-Dimensional Motion in a Gravitational Field

Given : v(0) = 〈u0, v0〉 & r(0) = 〈x0, y0〉

v(t) = 〈x′(t), y′(t)〉 = 〈u0,−gt+ v0〉

r(t) = 〈x(t), y(t)〉 =

⟨u0t+ x0,−

1

2gt2 + v0t+ y0

⟩

4

9.3 Two-Dimensional Motion

Given : v(0) = 〈|v0| cos θ, |v0| sin θ〉 & r(0) = 〈0, 0〉

Time =2|v0| sin θ

g

Range =|v0|2 sin 2θ

g

MaxHeight = y

(T

2

)=

(|v0| sin θ)2

2g

10 Planes and Surfaces

10.1 Plane Equations

The plane passing through the point P0(x0, y0, z0) with a normal vector n =〈a, b, c, 〉 is described by the equations:

a(x− x0) + b(y − y0) + c(z − z0) = 0

ax+ by + cz = d, where d = ax0 + by0 + cz0

In order to find the equation of a plane when given three points, simply createany two vectors out of the points and take the cross product to find the vectornormal to the plane. Then use one of the above formulae.

10.2 Parallel and Orthogonal Planes

Two planes are parallel if their normal vectors are parallel. Two planes areorthogonal if their normal vectors are orthogonal.

10.3 Surfaces

10.3.1 Ellipsoid

x2

a2+y2

b2+z2

c2= 1

10.3.2 Elliptic Paraboloid

z =x2

a2+y2

b2

It would be worth it to learn how to derive sections 9.2 and 9.3.

5

10.3.3 Hyperboloid of One Sheet

x2

a2+y2

b2− z2

c2= 1

10.3.4 Hyperboloid of Two Sheets

−x2

a2− y2

b2+z2

c2= 1

10.3.5 Elliptic Cone

x2

a2+y2

b2=z2

c2

10.3.6 Hyperbolic Paraboloid

z =x2

a2− y2

b2

11 Graphs and Level Curves

11.1 Functions of Two Variables

R2 → R

z = f(x, y)

F (x, y, z) = 0

11.2 Functions of Three Variables

R3 → R

w = f(x, y, z)

F (w, x, y, z) = 0

11.3 Level Curves

Imagine stepping onto a surface and walking along a path with constant eleva-tion. The path you walk on is known as the contour curve, while the projectionof the path onto the xy-plane is known as a level curve.

6

12 Limits and Continuity

12.1 Limits

The function f has the limit L as P (x, y) approaches P0(a, b).

lim(x,y)→(a,b)

f(x, y) = limP→P0

f(x, y) = L

If f(x, y) approaches two different values as (x, y) approaches (a, b) along twodifferent paths in the domain of f , then the limit does not exist.

12.2 Continuity

The function f if continuous at the point (a, b) provided:

lim(x,y)→(a,b)

f(x, y) = f(a, b)

13 Partial Derivatives

13.1 Definitions

fx(a, b) = limh→0

f(a+ h, b)− f(a, b)

h

fy(a, b) = limh→0

f(a, b+ h)− f(a, b)

h

So basically just take the derivative of one (the subscript) given that the otherone is a constant.

13.2 Notation for Higher-Order Partial Derivatives

∂

∂x

(∂f

∂x

)=∂2f

∂x2= (fx)x = fxx

∂

∂y

(∂f

∂y

)=∂2f

∂y2= (fy)y = fyy

∂

∂x

(∂f

∂y

)=

∂2f

∂x∂y= (fy)x = fyx

∂

∂y

(∂f

∂x

)=

∂2f

∂y∂x= (fx)y = fxy

Note: fxy = fyx for nice functions.

13.3 Differentiability

Suppose the function f has partial derivatives fx and fy defined on an openregion containing (a, b), with fx and fy continuous at (a, b). Then f is differen-tiable at (a, b). This also implies that it is continuous at (a, b).

7

14 Chain Rule

14.1 Examples

You can use a tree diagram to determine the equation for the chain rule. Youcan also just think about it. Refer to the following examples.

z is a function of x and y, while x and y are functions of t

dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt

w is a function of x, y, and z, while x, y, and z are functions of t

dw

dt=∂w

∂x

dx

dt+∂w

∂y

dy

dt+∂w

∂z

dz

dt

z is a function of x and y, while x and y are functions of s and t

∂z

∂s=∂z

∂x

∂x

∂s+∂z

∂y

∂y

∂s

w is a function of z, z is a function of x and y, x and y are functions of t

dw

dt=dw

dz

(∂z

∂x

dx

dt+∂z

∂y

dy

dt

)

14.2 Implicit Differentiation

Let F be differentiable on its domain and suppose that F (x, y) = 0 defines y asa differentiable function of x. Provided Fy 6= 0,

dy

dx= −Fx

Fy

15 Directional Derivatives and Gradient

15.1 Definitions

Let f be differentiable at (a, b) and let u = 〈u1, u2〉 be a unit vector in thexy-plane. The directional derivative of f at (a, b) in the direction of u is

Duf(a, b) = 〈fx(a, b), fy(a, b)〉 · 〈u1, u2〉 = ∇f(a, b) · u

Gradient∇f(x, y) = 〈fx(x, y), fy(x, y)〉 = fx(x, y)i + fy(x, y)j

8

15.2 Directions of Change

• f has its maximum rate of increase at (a, b) in the direction of the gradient∇f(a, b). The rate of increase in this direction is |∇f(a, b)|.

• f has its maximum rate of decrease at (a, b) in the direction of the gradient−∇f(a, b). The rate of decrease in this direction is −|∇f(a, b)|.

• The directional derivative is zero in any direction orthogonal to ∇f(a, b).

15.3 Expanding to Three Dimensions

It’s really intuitive how it expands into three dimensions. Just add anothercomponent or fz where you think it should go.

16 Tangent Plane and Linear Approximation

16.1 Tangent Plane for F(x,y, z) = 0

The tangent plane passes through the point P0(a, b, c).

Fx(a, b, c)(x− a) + Fy(a, b, c)(y − b) + Fz(a, b, c)(z − c) = 0

16.2 Tangent Plane for z = f(x,y)

The tangent plane passes through the point (a, b, f(a, b)).

z = fx(a, b)(x− a) + fy(a, b)(y − b) + f(a, b)

16.3 Linear Approximation

Firstly, calculate the equation of the tangent plane of a point near the point youwish to approximate. Then simply plug in the point and you’re done.

16.4 The differential dz

The change in z = f(x, y) as the independent variables change from (a, b) to(a+ dx, b+ dy) is denoted ∆z and is approximated by the differential dz:

∆z ≈ dz = fx(a, b)dx+ fy(a, b)dy

17 Max-Min Problems

17.1 Derivatives and Local Maximum/Minimum Values

If f has a local maximum or minimum value at (a, b) and the partial derivativesfx and fy exist at (a, b), then fx(a, b) = fy(a, b) = 0.

9

17.2 Critical Points

A critical point exists if either

• fx(a, b) = fy(a, b) = 0

• one (or both) of fx or fy does not exist at (a, b)

17.3 Second Derivative Test

Let D(x, y) = fxxfyy − f2xy

• If D(a, b) > 0 and fxx(a, b) < 0, then f has a local maximum at (a, b).

• If D(a, b) > 0 and fxx(a, b) > 0, then f has a local minimum at (a, b).

• If D(a, b) < 0, then f has a saddle point at (a, b).

• If D(a, b) = 0, then the test is inconclusive.

17.4 Absolute Maximum/Minimum Values

Let f be continuous on a closed bounded set R in R2. To find absolute maximumand minimum values of f on R:

1. Determine the values of f at all critical points in R.

2. Find the maximum and minimum values of f on the boundary of R.

3. The greatest function value found in Steps 1 and 2 is the absolute maxi-mum value of f on R, and the least function value found in Steps 1 and 2is the absolute minimum values of f on R.

18 Double Integrals

18.1 Double Integrals on Rectangular Regions

Let f be continuous on the rectangular region R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤d}. The double integral of f over R may be evaluated by either of two iteratedintegrals: ∫∫

R

f(x, y) dA =

∫ d

c

∫ b

a

f(x, y) dx dy =

∫ b

a

∫ d

c

f(x, y) dy dx

10

18.2 Double Integrals over Nonrectangular Regions

Let R be a region bounded below and above by the graphs of the continuousfunctions y = g(x) and y = h(x), respectively, and by the lines x = a and x = b.If f is continuous on R, then∫∫

R

f(x, y) dA =

∫ b

a

∫ h(x)

g(x)

f(x, y) dy dx

Let R be a region bounded on the left and right by the graphs of the continuousfunctions x = g(y) and x = h(y), respectively, and by the lines y = c and y = d.If f is continuous on R, then∫∫

R

f(x, y) dA =

∫ d

c

∫ h(y)

g(y)

f(x, y) dx dy

18.3 Areas of Regions by Double Integrals

area of R =

∫∫R

dA

19 Polar Double Integrals

19.1 Double Integrals over Polar Rectangular Regions

Let f be continuous on the region in the xy-plane R = {(r, θ) : 0 ≤ a ≤ r ≤b, α ≤ θ ≤ β}, where β − α ≤ 2π. Then∫∫

R

f(r, θ) dA =

∫ β

α

∫ b

a

f(r, θ) r dr dθ

19.2 Double Integrals over More General Polar Regions

Let f be continuous on the region in the xy-plane

R = {(r, θ) : 0 ≤ g(θ) ≤ r ≤ h(θ), α ≤ θ ≤ β}

where β − α ≤ 2π. Then.∫∫R

f(r, θ) dA =

∫ β

α

∫ h(θ)

g(θ)

f(r, θ) r dr dθ

If f is nonnegative on R, the double integral gives the volume of the solidbounded by the surface z = f(r, θ) and R.

11

19.3 Area of Polar Regions

A =

∫∫R

dA =

∫ β

α

∫ h(θ)

g(θ)

r dr dθ

20 Triple Integrals

Let D = {(x, y, z) : a ≤ x ≤ b, g(x) ≤ y ≤ h(x), G(x, y) ≤ z ≤ H(x, y)}, whereg, h, G, H are continuous functions. The triple integral of a continuous functionf on D is evaluated as the iterated integral∫∫∫

D

f(x, y, z) dV =

∫ b

a

∫ h(x)

g(x)

∫ H(x,y)

G(x,y)

f(x, y, z) dz dy dx

21 Cylindrical and Spherical Coordinates

21.1 Definitions

21.1.1 Cylindrical Coordinates

(r, θ, z) An extension of polar coordinates into R3. Simply add a z component.

21.1.2 Spherical Coordinates

(ρ, ϕ, θ)

• ρ is the distance from the origin to a point P .

• ϕ is the angle between the positive z-axis and the line OP .

• θ is the same angle as in cylindrical coordinates; it measure rotation aboutthe z-axis relative to the positive x-axis.

21.2 Rectangular to Cylindrical

r2 = x2 + y2

tan θ =y

xz = z

21.3 Cylindrical to Rectangular

x = r cos θ

y = r sin θ

z = z

12

21.4 Integration in Cylindrical Coordinates∫∫∫D

f(r, θ, z) dV =

∫ β

α

∫ h(θ)

g(θ)

∫ H(r cos θ,r sin θ)

G(r cos θ,r sin θ)

f(r, θ, z) dz r dr dθ

21.5 Rectangular to Spherical

ρ2 = x2 + y2 + z2

You have to solve for ϕ and θ with trigonometry.

21.6 Spherical to Rectangular

x = ρ sinϕ cos θ

y = ρ sinϕ sin θ

z = ρ cosϕ

21.7 Integration in Spherical Coordinates∫∫∫D

f(ρ, ϕ, θ) dV =

∫ β

α

∫ b

a

∫ h(ϕ,θ)

g(ϕ,θ)

f(ρ, ϕ, θ)ρ2 sinϕdρ dϕdθ

22 Change of Variables

22.1 Jacobian Determinant of a Transformation of TwoVariables

Given a transformation T : x = g(u, v), y = h(u, v), where g and h are differen-tiable on a region of the uv-plane, the Jacobian determinant of T is

J(u, v) =∂(x, y)

∂(u, v)=

∣∣∣∣∣∣∣∣∣∂x

∂u

∂x

∂v

∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣∣∣22.2 Change of Variables for Double Integrals∫∫

R

f(x, y) dA =

∫∫S

f(g(u, v), h(u, v))|J(u, v)| dA

22.3 Change of Variables for Triple Integrals

I am SO not typing out the expansion of the above into triple integrals. It’sintuitive. Just add stuff where you think it should go.

13

22.4 YOU have to Choose the Transformation

Just cry.

23 Vector Fields

23.1 Vector Fields in Two Dimensions

F(x, y) = 〈f(x, y), g(x, y)〉

23.2 Radial Vector Fields in R2

Let r = (x, y). A vector field of the form F = f(x, y)r, where f is a scalar-valuedfunction, is a radial vector field.

F(x, y) =r

|r|p=〈x, y〉|r|p

p is a real number. At every point (sans origin), the vectors of this field are

directed outward format he origin with a magnitude of |F| =1

|r|p−1. You can

also apply all of this to R3 by just adding a z component.

23.3 Gradient Fields and Potential Functions

Let z = ϕ(x, y) and w = ϕ(x, y, z) be differentiable functions on regions of R2

and R3, respectively. The vector field F = ∇ϕ is a gradient field, and thefunction ϕ is a potential function for F.

24 Line Integrals

24.1 Evaluating Scalar Line Integrals in R2

Let f be continuous on a region containing a smooth curve C : r(t) = 〈x(t), y(t)〉,for a ≤ t ≤ b. Then∫

C

f ds =

∫ b

a

f(x(t), y(t))|r′(t)| dt =

∫ b

a

f(x(t), y(t))√x′(t)2 + y′(t)2 dt

24.2 Evaluating Scalar Line Integrals in R3

Simply add a z component to the above where it obviously belongs.

14

24.3 Line Integrals of Vector Fields

24.3.1 Definition

Let F be a vector field that is continuous on a region containing a smoothoriented curve C parametrized by arc length. Let T be the unit tangent vectorat each point of C consistent with the orientation. The line integral of F overC is

∫C

F ·T ds.

24.3.2 Different Forms

F = 〈f, g, h〉 and C has a parametrization r(t) = 〈x(t), y(t), z(t)〉, for a ≤ t ≤ b∫ b

a

F ·r′(t) dt =

∫ b

a

(fx′(t)+gy′(t)+hz′(t)) dt =

∫C

f dx+g dy+h dz =

∫C

F ·dr

For line integrals in the plane, we let F = 〈f, g〉 and assume C is parametrizedin the form r(t) = 〈x(t), y(t)〉, for a ≤ t ≤ b. Then∫

C

F ·T ds =

∫ b

a

(fx′(t) + gy′(t)) dt =

∫C

f dx+ g dy =

∫C

F · dr

24.4 Work

F is a force field

W =

∫C

F ·T ds =

∫ b

a

F · r′(t) dt

24.5 Circulation

F is a vector field

Circulation =

∫C

F ·T ds

24.6 Flux

Flux =

∫C

F · n ds =

∫ b

a

(fy′(t)− gx′(t)) dt

n = T× k, and a positive answer means a positive outward flux.

25 Conservative Vector Fields

25.1 Test for Conservative Vector Field

Let F = 〈f, g, h〉 be a vector field defined on a connected and simply connectedregion D of R3, where f , g, and h have continuous first partial derivatives on

15

D. Then, F is a conservative vector field on D (there is a potential function ϕsuch that F = ∇ϕ) if and only if

• ∂f

∂y=∂g

∂x

• ∂f

∂z=∂h

∂x

• ∂g

∂z=∂h

∂y

For vector fields in R2, we have the single condition∂f

∂y=∂g

∂x.

25.2 Finding Potential Functions

Suppose F = 〈f, g, h〉 is a conservative vector field. To find ϕ such that F = ∇ϕ,take the following steps:

1. Integrate ϕx = f with respect to x to obtain ϕ, which includes an arbitraryfunction c(y, z.

2. Compute ϕy and equate it to g to obtain an expression for cy(y, z).

3. Integrate cy(y, z) with respect to y to obtain c(y, z), including an arbitraryfunction d(z).

4. Compute ϕz and equate it to h to get d(z).

Beginning the procedure with ϕy = g or ϕz = h may be easier in some cases.This method can also be used to check if a vector field is conservative by seeingif there is a potential function.

25.3 Fundamental Theorem for Line Integrals∫C

F ·T ds =

∫C

F · dr = ϕ(B)− ϕ(A)

25.4 Line Integrals on Closed Curves

Let R in R2 (or D in R3) be an open region. Then F is a conservative vectorfield on R if and only if

∮C

F ·dr = 0 on all simple closed smooth oriented curvesC in R.

26 Green’s Theorem

26.1 Circulation Form∮C

F · dr =

∮C

f dx+ g dy =

∫∫R

(∂g

∂x− ∂f

∂y

)dA

16

26.2 Area of a Plane Region by Line Integrals∮C

x dy = −∮C

y dx =1

2

∮C

(x dy − y dx)

26.3 Flux Form∮C

F · n ds =

∮C

f dy − g dx =

∫∫R

(∂f

∂x+∂g

∂y

)dA

27 Divergence and Curl

27.1 Divergence of a Vector Field

div(F) = ∇ · F =∂f

∂x+∂g

∂y+∂h

∂z

27.2 Divergence of Radial Vector Fields

div(F) =3− p|r|p

F =r

|r|p=

〈x, y, z〉(x2 + y2 + z2)p/2

27.3 Curl

curl(F) = ∇× F

Just derive the curl by doing the cross product.

27.4 Divergence of the Curl

∇ · (∇× F) = 0

28 Surface Integrals

28.1 Parameterization

28.1.1 z is Explicitly Defined

Use x = x, y = y, and since z is explicitly defined, you already have what zequals.

28.1.2 Cylinder

Simply use cylindrical coordinates to parameterize the surface in terms of θ andz.

17

28.1.3 Sphere

Simply use spherical coordinates to parameterize the surface in terms of ϕ andθ.

28.1.4 Cone

Use:

• x = v cosu

• y = v sinu

• z = v

0 ≤ u ≤ 2π and 0 ≤ v ≤ h

28.2 Surface Integrals of Parameterized Surfaces∫∫Σ

f(x, y, z) dσ =

∫∫R

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

∣∣∣∣∂r

∂s× ∂r

∂t

∣∣∣∣ dA29 Divergence Theorem

Let F be a vector field whose components have continuous first partial deriva-tives in a connected and simply connected region D enclosed by a smooth ori-ented surface S. Then ∫∫

S

F · n dS =

∫∫∫D

∇ · F dV

where n is the outward normal vector on S.

18