final review ii

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Final Review Session II

Math 21a

May 8, 2008

.

.Image: Flickr user Rileyroxx

http://flickr.com/photos/rileyroxx/162970439/

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Announcements

◮ Review sessions:◮ Tuesday, May 6 - Hall D 4-5:30pm◮ Thursday, May 8 - Hall A 4-5:30pm◮ Monday, May 12 - Hall C 5-6:30pm

◮ Final Exam: 5/23 9:15 am, Emerson 105 (tentative)◮ Office hours during reading period on website

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Part I

Chapter 11: Partial Derivatives

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Outline

Functions of Several Variables

Limits and Continuity

Partial DerivativesDefinitionClairaut’s Theorem

Tangent Planes and Linear Approximations

The Chain RuleThe Chain RuleImplicit Differentiation

Directional Derivatives and the Gradient Vector

Maximum and Minimum Values

Lagrange Multipliers

. . . . . .

Functions of Several VariablesLearning Objectives

◮ To understand functions of several variables and be able torepresent these functions using level sets.

. . . . . .

Drawing level sets and contour plots

◮ The level curves of a function of two variables are the curvef(x, y) = C, for various values of C.

◮ To draw a level curve: try to make it look like◮ the graph of a function◮ or a conic section

. . . . . .

Example (Chapter 11 Review, #6)Sketch several level curves of the function

f(x, y) = ex + y

SolutionIf ex + y = C, then y = C − ex.So the level curves look liketranslated mirror images of thestandard exponential graph.

..x

.y

.−2

.−1

.0

.1

.2

.3

. . . . . .

Outline









. . . . . .

Limits and ContinuityLearning Objectives

◮ To understand and be able to apply the concept of a limit of afunction of several variables.

◮ To understand and be able to apply the definition of continuityfor a function of several variables.

. . . . . .

Outline









. . . . . .

Partial DerivativesLearning Objectives

◮ To understand and be able to apply the definition of a partialderivative.

◮ To be able to compute partial derivatives.◮ To understand and be able to apply Clairaut’s Theorem. If f is

defined on a disk D that contains the point (a, b) and thefunctions fxy and fyx are continuous on D, then

fxy(a, b) = fyx(a, b).

◮ To understand the idea of a partial differential equation, and tobe able to verify solutions to partial differential equations.

. . . . . .

Definition

DefinitionLet f(x, y) be a function of two variables. We define the partial

derivatives∂f∂x

and∂f∂y

at a point (a, b) as

∂f∂x

(a, b) = limh→0

f (a + h, b) − f (a, b)h

∂f∂y

(a, b) = limh→0

f (a, b + h) − f (a, b)h

In other words, we temporarily treat the other variable as constantand differentiate the resulting one-variable function as in Calculus I.

. . . . . .

Example (Chapter 11 Review, #16)Find all the partial derivatives of w =

xy − z

Solution

∂w∂x

=1

y − z∂w∂y

= − x(y − z)2

∂w∂z

=x

(y − z)2

. . . . . .

Second derivatives

If f(x, y) is a function of two variables, each of its partial derivativesare function of two variables, and we can hope that they aredifferentiable, too. So we define the second partial derivatives.

∂2f∂x2 =

∂

∂x

(∂f∂x

)= fxx

∂2f∂y ∂x

=∂

∂y

(∂f∂x

)= fxy

∂2f∂x ∂y

=∂

∂x

(∂f∂y

)= fyx

∂2f∂y2 =

∂

∂y

(∂f∂y

)= fyy

. . . . . .

Example (Chapter 11 Review, #22)Find all second partial derivatives of v = r cos(s + 2t).

SolutionWe have

vr = cos(s + 2t) vs = −r sin(s + 2t) vt = −2r sin(s + 2t).

So

vrr = 0 vsr = − sin(s + 2t) vtr = −2 sin(s + 2t)

vrs = − sin(s + 2t) vss = −r cos(s + 2t) vts = −2r cos(s + 2t)

vrt = −2 sin(s + 2t) vst = −2r cos(s + 2t) vtt = −4r cos(s + 2t)

. . . . . .

Clairaut’s Theorem

The “mixed partials” bookkeeping may seem scary. However, we aresaved by:

Theorem (Clairaut’s Theorem/Young’s Theorem)If f is defined near (a, b) and fxy and fyx are continuous at (a, b), then

fxy(a, b) = fyx(a, b).

The upshot is that we needn’t worry about the ordering.

. . . . . .

Example (Chapter 11 Review, #22)Find all second partial derivatives of v = r cos(s + 2t).

SolutionWe have

vr = cos(s + 2t) vs = −r sin(s + 2t) vt = −2r sin(s + 2t).

So

vrr = 0 vsr = − sin(s + 2t) vtr = −2 sin(s + 2t)

vrs = − sin(s + 2t) vss = −r cos(s + 2t) vts = −2r cos(s + 2t)

vrt = −2 sin(s + 2t) vst = −2r cos(s + 2t) vtt = −4r cos(s + 2t)

. . . . . .

Partial Differential Equations

DefinitionA partial differential equation (PDE) is a differential equationfor a function of more than one variable. So the derivatives involvedare partial derivatives.

. . . . . .

Example (Chapter 11 Review, #24)If z = sin(x + sin t), show that

∂z∂x

∂2z∂x ∂t

=∂z∂t

∂2z∂x2

We simply take the derivatives:

∂z∂x

= cos(x + sin t)∂z∂t

= cos(x + sin t) cos t

∂2z∂x ∂t

= − sin(x + sin t) cos t∂2z∂t2

= − sin(x + sin t)

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Outline









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Tangent Planes and Linear Approximations ILearning Objectives

◮ To understand the concept of a tangent plane to a surfacez = f(x, y) and to be able to compute the equation of tangentplanes

◮ To understand and be able to find a linear approximation to afunction z = f(x, y).

◮ To understand the concept of a tangent plane to a parametricallydefined surface

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

and to be able to compute the equation of tangent planes.

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Tangent Planes and Linear Approximations IILearning Objectives

◮ To understand and be able to find the differential to a functionz = f(x, y),

dz =∂z∂x

dx +∂z∂y

dy

To be able to use the differential to estimate maximum error.

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Fact

◮ Suppose f has continuous partial derivatives. An equation of thetangent plane to the surface z = f(x, y) at the point P(x0, y0, z0) is

z − z0 =∂f∂x

(x0, y0)(x − x0) +∂f∂y

(x0, y0)(y − y0)

◮ This is also the best linear approximation to the function f near(x0, y0).

. . . . . .

Example (Chapter 11 Review, #28)Find equations for the tangent plane and normal line to the surfacexy + yz + xz = 3 at the point (1, 1, 1).

SolutionSolve the equation for z as a function of x: z =

3 − xyx + y

. So

zx = − y2 + 3(x + y)2 zy = − x2 + 3

(x + y)2

zx(1, 1) = −1 zy(1, 1) = −1

Thus an equation for the tangent plane is

z = 1 − (x − 1) − (y − 1) =⇒ x + y + z = 3

. . . . . .

Solution (Normal line)If the plane has equation x + y + z = 3, then n = ⟨1, 1, 1⟩ is normal tothe plane. So the system

x − 1 = y − 1 = z − 1 =⇒ x = y = z

describes the normal line.

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Outline









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The Chain RuleLearning Objectives

◮ To understand and be able to apply the chain rule for functionsof several variables.

◮ To understand and be able to implicitly differentiate functions.

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Fact (The Chain Rule, version I)When z = F(x, y) with x = f(t) and y = g(t), then

z′(t) = Fx(f(t), g(t))f′(t) + Fy(f(t), g(t))g′(t)

or

dzdt

=∂F∂x

dxdt

+∂F∂y

dydt

We can generalize to more variables, too. If F is a function ofx1, x2, . . . , xn, and each xi is a function of t, then

dzdt

=∂F∂x1

dx1

dt+

∂F∂x2

dx2

dt+ · · · + ∂F

∂xn

dxn

dt

. . . . . .

Tree Diagrams for the Chain Rule

..F

.x

.t

.dxdt

.∂F∂x

.y

.t

.dydt

.∂F∂y

To differentiate with respect to t, find all “leaves” marked t. Goingdown each branch, chain (multiply) all the derivatives together. Thenadd up the result from each branch.

dzdt

=dFdt

=∂F∂x

dxdt

+∂F∂y

dydt

. . . . . .

Example (Chapter 11 Review, #40)The length x of a side of a triangle is increasing at a rate of 3 in/s, thelength y of another side is increasing at the rate of 2 in/s, and thecontained angle θ is increasing at the rate of 0.05 radian/s. How fast isthe area of the triangle changing when x = 40 in, y = 50 in, θ = π/6?

SolutionWe have A = 1

2xy sin θ, so

dAdt

=∂A∂x

dxdt

+∂A∂y

dydt

+∂A∂θ

dθdt

=12

y sin(θ)x′ +12

x sin(θ)y′ +12

xy cos(θ)θ′

. . . . . .

Solution (continued)At our point in time we have

A′ = 12(50)

( 12

)(3) + 1

2(40)( 1

2

)(2) + 1

2(40)(50)

(√3

2

) (120

)=

352

+ 25√

3

. . . . . .

Fact (The Chain Rule, Version II)When z = F(x, y) with x = f(t, s) and y = g(t, s), then

∂z∂t

=∂F∂x

∂x∂t

+∂F∂y

∂y∂t

∂z∂s

=∂F∂x

∂x∂s

+∂F∂y

∂y∂s

..F

.x

.t .s

.y

.t .s

. . . . . .

Example (Chapter 11 Review, #36)If v = x2 sin y + yexy, where x = s + 2t and y = st, use the Chain Ruleto find ∂v/∂s and ∂v/∂t when s = 0 and t = 1.

SolutionWe have

∂v∂s

=∂v∂x

∂x∂s

+∂v∂y

∂y∂s

= (2x sin y + y2exy)(1) + (x2 cos y + exy + xyexy)(t)

When s = 0 and t = 1, then y = 0 and x = 2. So vs(0, 1) = 5.

. . . . . .

Solution (Continued)Likewise

∂v∂t

=∂v∂x

∂x∂t

+∂v∂y

∂y∂t

= (2x sin y + y2exy)(2) + (x2 cos y + exy + xyexy)(s)

When s = 0 and t = 1, then y = 0 and x = 2. So vt(0, 1) = 0.

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Implicit Differentiation

FactAlong the level curve F(x, y) = c, the slope of the tangent line is given by

dydx

=

(dydx

)F= −∂F/∂x

∂F/∂x= −Fx(x, y)

Fy(x, y)

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Example (Chapter 11 Review, #42)

If yz4 + x2z3 = exyz, find∂z∂x

and∂z∂y

Solution (Chain Rule). Treating z implicitly as a function of x and y we have

y(

4z∂z∂x

)+ 2xz3 + x2

(3z2 ∂z

∂x

)= exyz

(yz + xy

∂z∂x

)=⇒ ∂z

∂x=

yzexyz − 2xz3

4yz3 + 3x2z2 − xyexyz

Similarly∂z∂y

=xzexyz − z4


. . . . . .

Solution (with differentials)Taking the total differential of yz4 + x2z3 = exyz gives

z4 dy + 4yz3 dz + 2xz3dx + 3x2z2 dz = exyz(yz dx + xz dy + xy dz)

To find ∂z/∂x, set dy = 0 and solve for dz/dx:

∂z∂x

=

(dzdx

)dy=0

=yzexyz − 2xz3


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Outline









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Directional Derivatives and the Gradient VectorLearning Objectives

◮ To understand and be able to apply the definition of thedirectional derivative.

◮ To understand and be able to apply the definition of the gradientof a function f.

◮ To understand that |∇f(x)| is the maximum value of thedirectional derivative, Duf(x).

◮ To be able to use the gradient to find the tangent plane to alevel surface.

. . . . . .

DefinitionThe directional derivative of f at (x0, y0) in the direction of aunit vector u = ⟨a, b⟩ is

Duf(x0, y0) = limh→0

f(x0 + ha, y0 + hb) − f(x0, y0)

h

if this limit exists.

DefinitionIf f is a function of two variables x and y, then the gradient of f isthe vector field grad f = ∇f defined by

∇f(x, y) =

⟨∂f∂x

(x, y),∂f∂y

(x, y)⟩

=∂f∂x

(x, y)i +∂f∂x

(x, y)j

. . . . . .

Fact

Duf(x, y) = ∇f(x, y) · u.

. . . . . .


(a) When is the directional derivative of f a maximum?

(b) When is it a minimum?

(c) When is it 0?

(d) When is it half of its maximum value?

. . . . . .

SolutionNotice that

Duf(x, y) = ∇f(x, y) · u = |∇f(x, y)| · |u| cos θ = |∇f(x, y)| cos θ

where θ is the angle between ∇f(x, y) and u.

(a) It is maximal when u and ∇f point in the same direction

(b) It is minimal when u and ∇f point in opposite directions

(c) It is 0 when u and ∇f are perpendicular

(d) It is half its maximum value when θ = π/3 (so cos θ = 1/2)

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Outline









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Maximum and Minimum ValuesLearning Objectives

◮ To understand and be able to find the local extrema and saddlepoints of a function z = f(x, y).

◮ To understand and be able to apply the Second Derivative Test.◮ To understand and be able to find the absolute maximum and

minimum of a function z = f(x, y).

. . . . . .

Theorem (The Second Derivative Test)Suppose the second partial derivatives of f are continuous on a disk withcenter (a, b) and suppose that fx(a, b) = 0 and fy(a, b) = 0 [that is,(a, b) is a critical point of f]. Let

D = D(a, b) = fxx(a, b)fyy(a, b) −[fxy(a, b)

]2=

∣∣∣∣fxx(a, b) fyx(a, b)fxy(a, b) fyy(a, b)

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

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

(c) If D < 0, then (a, b) is a saddle point of f

(d) (If D = 0, the second derivative test says nothing.)

. . . . . .

Example (Chapter 11 Review, #56)Find the maximum and minimum values of f(x, y) = e−x2−y2

(x2 + 2y2).

SolutionNotice that

fx = −2x(x2 + 2y2 − 1)e−x2−y2

fy = −2y(x2 + 2y2 − 2)e−x2−y2

So the critical points are (0, 0), (0, 1), (0,−1), (1, 0), (−1, 0).

. . . . . .

Solution (Continued)The second derivatives are

fxx = 2e−x2−y2(

2x4 + 4y2x2 − 5x2 − 2y2 + 1)

fxy = 4e−x2−y2xy

(x2 + 2y2 − 3

)fyy = 2e−x2−y2

(4y4 + 2x2y2 − 10y2 − x2 + 2

)Thus

D(0, 0) = 8 D(±1, 0) = −8e−2 D(0,±1) = 16e−2

fxx(0, 0) = 2 fyy(0,±1) = −2e−1

This means f(0, 0) = 0 is a local minimum, f(0,±1) = 2e−1 are localmaxima, and (±1, 0) are saddle points.

. . . . . .

-2 -1 0 1 2

-2

-1

0

1

2

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Outline









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Lagrange MultipliersLearning Objectives

◮ To understand and be able to apply the Method of LagrangeMultipliers to solve constrained optimization problems. To findthe minimum and maximum values of f(x, y, z) subject to theconstraint g(x, y, z) = k,

1. Find all values of x, y, z, λ such that

∇f(x, y, z) = λ∇g(x, y, z)

and g(x, y, z) = k.2. Evaluate f at all of the points (x, y, z) from step (1). The largest of

these values will be the maximum value of f and the smallest theminimum value of f.

. . . . . .


(x2 + 2y2)subject to the constraint that x2 + y2 = 4,

SolutionWe have

2λx = fx = −2x(x2 + 2y2 − 1)e−x2−y2= −2x(y2 + 3)e−x2−y2

2λy = fy = −2y(x2 + 2y2 − 2)e−x2−y2= −2x(y2 + 2)e−x2−y2

If x ̸= 0 and y ̸= 0, then canceling the both gives a contradiction. Sox = 0, giving the points (0,±2), or y = 0, giving (±2, 0). Since

f(±2, 0) = 4e−4 f(0,±2) = 8e−4

The maximum value of the constrained function is 8e−4, and the minimumvalue is 4e−4.

. . . . . .


(x2 + 2y2)on the set x2 + y2 ≤ 4.

Solution

◮ The unconstrained extrema are 0 and 2e−1.◮ The constrained extrema are 8e−4 and 4e−4.

The largest of these values is 2e−1, and the smallest is 0.

. . . . . .

Example (Chapter 11 Review, #64)A package in the shape of a rectangular box can be mailed by the U.S.Postal Service if the sum of its length and girth (the perimeter of across-section perpendicular to the length) is at most 108 in. Find thedimensions of the package with the largest volume that can be mailed.

SolutionThe problem is to maximize f(x, y, z) = xyz subject to the constraint that

g(x, y, z) = x + 2y + 2z = 108 in3

. . . . . .

Solution (continued)The lagrange multiplier equations are

yz = λ xz = 2λ xy = 2λ

Multiply each equation by x, y, and z respectively and add. We get3xyz = 108λ. Divide each of the above into this and we get

x = 36 in y = 18 in z = 18 in

So the critical point is f(36, 18, 18) = 11, 664 in3.

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Part II

Chapter 12: Multiple Integrals

. . . . . .

Outline

Double Integrals over Rectangles; Iterated Integrals

Double Integrals over General Regions

Double Integrals in Polar Coordinates

Surface AreaGraphsSurfaces of Revolution

Triple Integrals

Triple Integrals in Cylindrical and Spherical Coordinates

. . . . . .

Double Integrals over Rectangles; Iterated Integrals ILearning Objectives

◮ To understand and be able to apply the definition of the doubleintegral of a function f over a region R,∫∫

R

f(x, y) dA.

◮ To understand and be able to apply the midpoint rule for thedouble integral of a function f over a region R.

. . . . . .

Double Integrals over Rectangles; Iterated Integrals IILearning Objectives

◮ To understand and to be able to compute the value of∫∫R

f(x, y) dA.

when R is a rectangle in the xy-plane. In particular, to be able toevaluate the integral as an iterated integral and to be able toapply Fubini’s Theorem.

◮ To understand and be able to apply the properties of doubleintegrals.

◮ To understand and to be able to compute the average value of afunction z = f(x, y) over a region R in the xy-plane.

. . . . . .

DefinitionFor each m and n, divide the interval [a, b] into m subintervals ofequal width, and the interval [c, d] into n subintervals. For each i andj, form the subrectangles

Rij = [xi−1, xi] × [yj−1, yj]

Choose a sample point (x∗ij , y∗ij ) in each subrectangle and form theRiemann sum. The double integral of f over the rectangle R is∫∫

R

f(x, y) dA = limm,n→∞

m∑i=1

n∑j=1

f(x∗ij , y∗ij )∆y ∆x

FactFor continuous f this limit is the same regardless of method for choosingthe sample points.

. . . . . .

Properties of Double Integrals I

(a)∫∫D

[f(x, y) + g(x, y)] dA =

∫∫D

f(x, y) dA +

∫∫D

g(x, y) dA

(b)∫∫D

cf(x, y) dA = c∫∫D

f(x, y) dA

(c) If f(x, y) ≥ g(x, y) for all (x, y) ∈ D, then∫∫D

f(x, y) dA ≥∫∫D

g(x, y) dA.

(d) If D = D1 ∪ D2, where D1 and D2 do not overlap except possiblyon their boundaries, then∫∫

D

[f(x, y)] dA =

∫∫D1

f(x, y) dA +

∫∫D2

f(x, y) dA

. . . . . .

Properties of Double Integrals II

(e)∫∫D

dA is the area of D, A(D).

(f) If m ≤ f(x, y) ≥ M for all (x, y) ∈ D, then

m · A(D) ≤∫∫D

f(x, y) dA ≤ M · A(D).

. . . . . .

Fubini’s Theorem

Double integrals look hard. Iterated integrals look easy/easier. Thegood news is:

Theorem (Fubini’s Theorem)If f is continuous on R = [a, b] × [c, d], then∫∫

R

f(x, y) dA =

∫ b

a

∫ d

cf(x, y) dy dx =

∫ d

c

∫ b

af(x, y) dx dy

This is also true if f is bounded on R, f is discontinuous only on a finitenumber of smooth curves, and the iterated integrals exist.

. . . . . .


Compute the value of the iterated integral∫ 1

0

∫ 1

0yexy dx dy

Solution

∫ 1

0

∫ 1

0yexy dx dy =

∫ 1

0[exy]x=1

x=0 dy

=

∫ 1

0(ey − 1) dy = e − 2

This is the same as the integral of f over the rectangle D = [0, 1] × [0, 1].

. . . . . .

Outline





Triple Integrals


. . . . . .

Double Integrals over General Regions ILearning Objectives

◮ To understand and be able to compute the double integral of afunction f over a type I region R,∫∫

D

f(x, y) dA =

∫ b

a

∫ g2(x)

g1(x)f(x, y) dy dx,

where

D = {(x, y)|a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}.

. . . . . .

Double Integrals over General Regions IILearning Objectives

◮ To understand and be able to compute the double integral of afunction f over a type II region R,∫∫

D

f(x, y) dA =

∫ d

c

∫ h2(y)

h1(y)f(x, y) dx dy,

where

D = {(x, y)|c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)}.

. . . . . .

Double Integrals over General Regions IIILearning Objectives

◮ To understand and be able to compute the double integral of afunction f over a general region D,∫∫

D

f(x, y) dA.

. . . . . .



0

∫ ex

x3xy2 dy dx

Solution

∫ 1

0

∫ ex

x3xy2 dy dx =

∫ 1

0

[xy3

]y=ex

y=xdx

=

∫ 1

0

(xe3x − x4

)dx

=29

e3 − 425 .

.x

.y

. . . . . .


Write∫∫

Rf(x, y) dA as an

iterated integral over theregion R shown.

..x

.y

.−4 .4

.4

SolutionWriting R as two regions of type I, we get∫ 0

−4

∫ x+4

0f(x, y) dy dx +

∫ 4

0

∫ 4−x

0f(x, y) dy dx

As a region of type II, we get∫ 4

0

∫ 4−y

y−4f(x, y) dx dy.

. . . . . .


Find∫∫

Dy dA, where D is the

region above the hyperbolaxy = 1 and the line y = x andbelow y = 2.

.

SolutionThe region is of type II. The iterated integral is∫ 2

1

∫ y

1/yy dx dy = 2

. . . . . .

Outline





Triple Integrals


. . . . . .

Double Integrals in Polar CoordinatesLearning Objectives

◮ To understand and be able to change to polar coordinates in adouble integral,∫∫

D

f(x, y) dA =

∫ β

α

∫ b

af(r cos θ, r sin θ) r dr dθ.

. . . . . .

Arc length and sector area

s = rθ

A =12

r2θ

(if θ is in radians).

.r

.θ

.s

. . . . . .

Area of a polar rectangle

∆A =12

r22 (θ2 − θ1) −12

r21 (θ2 − θ1)

=12

(r22 − r21

)(θ2 − θ1)

=12

(r2 + r1) (r2 − r1) (θ2 − θ1)

= r̄ ∆r ∆θ. .

.r1.

.r2. .θ1

..θ2

.̄r

.∆θ

. . . . . .

FactIf f is continuous on a polar region of the form

D = { (r, θ) | α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ) }

then ∫∫D

f(x, y) dA =

∫ β

α

∫ h2(θ)

h1(θ)f(r cos θ, r sin θ) r dr dθ.

◮ Notice the “area element” is r dr dθ, not dr dθ!

. . . . . .


Find∫∫

Dx dA, where D is the

region in the first quadrantthat lies between the circlesx2 + y2 = 1 and x2 + y2 = 2.

..x

.y

Solution

∫∫D

x dA =

∫ 2π

0

∫ √2

1r cos(θ) r dr dθ =

∫ 2π

0cos(θ) dθ ·

∫ √2

1r2 dr = 0

. . . . . .

Outline





Triple Integrals


. . . . . .

Surface AreaLearning Objectives

◮ To understand and be able to compute the surface area of aparameterized surface

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

where (u, v) are in a domain D.

A(S) =

∫∫D

|ru × rv| dA,

where

ru =∂x∂u

i +∂y∂u

j +∂z∂u

k rv =∂x∂v

i +∂y∂v

j +∂z∂v

k

. . . . . .

Surface AreaIILearning Objectives

◮ To understand and be able to compute the surface area of thegraph of a function z = f(x, y),

A(S) =

∫∫D

√1 +

(∂z∂x

)2

+

(∂z∂y

)2

dA.

◮ To understand and be able to compute the surface area of asurface of revolution S obtained by rotating a curve y = f(x),a ≤ x ≤ b, about the x-axis, where f(x) ≥ 0 and f′ is continuous,

A(S) = 2π

∫ b

af(x)

√1 + [f′(x)]2 dx.

. . . . . .

Parametrizing a surface

A parametric surface S is defined by a vector-valued function of twoparameters

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

where (u, v) varies throughout a region D in the plane.

..u

.v

.D .r

.x .y

.z

.S

. . . . . .

ApproximateThe vectors representing the sides of a small rectangle “pushforward” to vectors tangent to the surface.

..u

.v

.∆u i

.∆v j.D

.x .y

.z

.∆u ru.∆v rv .S

They span a parallelogram approximating this piece of the surface. Sothe area of that piece is

∆A ≈ |∆u ru × ∆v rv| = |ru × rv| ∆v ∆u

. . . . . .

Take the Limit

So

A ≈m∑

i=1

n∑j=1

∣∣ru(uij, vij) × rv(uij, vij)∣∣ ∆v ∆u

Taking the limit we get

A =

∫ b

a

∫ d

c|ru × rv| dv du

More generally, if D is any region (not necessarily a rectangle), thesurface area is ∫∫

D

|ru × rv| dA

Remembering the dA is over the domain (u, v) space.

. . . . . .

Example (Chapter 12 Review, #38(a))Set up, but don’t evaluate, an integral for the surface area of theparametric surface given by the vector functionr(u, v) = v2i − uvj + u2k, 0 ≤ u ≤ 3, −3 ≤ v ≤ 3.

SolutionWe have

ru = ⟨0,−v, 2u⟩ rv = ⟨2v,−u, 0⟩

ru × rv =⟨−2u2,−4uv,−2v2

⟩|ru × rv| = 2

√u4 + 4v2u2 + v4

So

S =

∫ 3

0

∫ 3

−32√

u4 + 4v2u2 + v4 dv du

. . . . . .

0

2

4

6

8

-5

0

5

0

2

4

6

8

. . . . . .

Surface areas of GraphsNotation as in Stewart, p. 869ff.

Suppose f(x, y) is a function oftwo variables with domain D.The graph of f over D is thesurface in space given by

S = { (x, y, z) | (x, y) ∈ D, z = f(x, y) } .

.D.x

.y

.z

.S

. . . . . .

To find the surface area of S, use the parametrization

r(x, y) = ⟨x, y, f(x, y)⟩

Then

rx =

⟨1, 0,

∂f∂x

⟩ry =

⟨0, 1,

∂f∂y

⟩rx × ry =

⟨− ∂f

∂x,− ∂f

∂y, 1

⟩So

A =

∫∫D

√1 +

(∂f∂x

)2

+

(∂f∂y

)2

dA

. . . . . .

Surfaces of Revolution

A surface of revolution can be described by rotating the graph ofy = f(x) over the interval [a, b] around the x-axis.

..x

.y

.z

. . . . . .

Choose the parametrization

r(x, θ) = ⟨u, f(u) cos θ, f(u) sin θ⟩

where a ≤ x ≤ b, 0 ≤ θ ≤ 2π. Then

rx =⟨1, f′(x) cos θ, f′(x) sin θ

⟩rθ = ⟨0,−f(x) sin θ, f(x) cos θ⟩

rx × rθ =⟨f′(x)f(x),−f(x) cos θ,−f(x) sin θ

⟩|ru × rv| = f(x)

√1 + f′(x)2

So

A =

∫ 2π

0

∫ b

af(x)

√1 + f′(x)2 dx dθ = 2π

∫ b

af(x)

√1 + f′(x)2 dx

. . . . . .

Outline





Triple Integrals


. . . . . .

Triple Integrals ILearning Objectives

◮ To understand and be able to apply the definition of the tripleintegral of a function f over a box B,

∫∫∫B

f(x, y, z) dV = liml,m,n→∞

l∑i=1

m∑j=1

n∑k=1

f(xi, yj, zk)∆V.

. . . . . .

Triple Integrals IILearning Objectives

◮ To understand and to be able to compute the value of∫∫∫B

f(x, y, z) dV

when B = [a, b] × [c, d] × [r, s]. In particular, to be able toevaluate the integral as an iterated integral and to be able toapply Fubini’s Theorem,∫∫∫

B

f(x, y, z) dV =

∫ s

r

∫ d

c

∫ b

af(x, y, z) dx dy dz.

. . . . . .

Triple Integrals IIILearning Objectives

◮ To understand and to be able to compute the value of∫∫∫E

f(x, y, z) dV =

∫ b

a

∫ g2(x)

g1(x)

∫ u2(x,y)

u1(x,y)f(x, y, z) dz dy dx

when

E = {a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x), u1(x, y) ≤ z ≤ u2(x, y)}.

. . . . . .

Triple Integrals IVLearning Objectives

◮ To understand and to be able to compute the value of∫∫∫E

f(x, y, z) dV =

∫ d

c

∫ h2(y)

h1(y)

∫ u2(x,y)

u1(x,y)f(x, y, z) dz dx dy

when

E = {c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y), u1(x, y) ≤ z ≤ u2(x, y)}.

◮ To be able to compute the volume of a region E by

V(E) =

∫∫∫E

dV.

. . . . . .



0

∫ y

0

∫ 1

x6xyz dz dx dy

Solution

∫ 1

0

∫ y

0

∫ 1

x6xyz dz dx dy =

∫ 1

0

∫ y

0

[3xyz2

]z=1

z=xdx dy

=

∫ 1

0

∫ y

0

(3xy − 3x3y

)dx dy

=

∫ 1

0

[32

x2y − 34

x4y]x=y

x=0dy

=

∫ 1

0

(32

y3 − 34

y5)

dy =14

. . . . . .

0.0

0.5

1.0

0.0

0.5

1.00.0

0.5

1.0

. . . . . .

Example (Chapter 12 Review, #48)Give five other iterated integrals that are equal to∫ 2

0

∫ y3

0

∫ y2

0f(x, y, z) dz dx dy

0

2

4

6

80.0

0.51.0

1.52.0

0

1

2

3

4

. . . . . .

Views of the Region

0

2

4

6

80.0

0.51.0

1.52.0

0

1

2

3

4

.

.8 .x

.2 .y

.Top .D1

.

.Front .D2

.2.y

.4.z

.

.Left .D3.D′

3

.D′′3

.8.x

.4.z

. . . . . .

From the top

.

.8 .x

.2 .y

.D1

D1 ={

0 ≤ y ≤ 2, 0 ≤ x ≤ y3}

={

0 ≤ x ≤ 8, x1/3 ≤ x ≤ y}

So

I =

∫∫D1

∫ y2

0f(x, y, z) dz dAx,y

=

∫ 2

0

∫ y3

0

∫ y2

0f(x, y, z) dz dx dy

=

∫ 8

0

∫ 2

x1/3

∫ y2

0f(x, y, z) dz dy dx

. . . . . .

From the front

.

.D2

.2.y

.4.z D2 =

{0 ≤ y ≤ 2, 0 ≤ z ≤ y2

}=

{0 ≤ z ≤ 4,

√z ≤ y ≤ 2

}So

I =

∫∫D2

∫ y3

0f(x, y, z) dy dAy,z

=

∫ 2

0

∫ y2

0

∫ y3

0f(x, y, z) dx dz dy

=

∫ 4

0

∫ 2

√z

∫ y3

0f(x, y, z) dx dy dz

. . . . . .

From the side

..D′

3

.D′′3

.x = y3

.z = y2

.z = x2/3

.8.x

.4.z D′

3 ={

0 ≤ x ≤ 8, 0 ≤ z ≤ x2/3}

={

0 ≤ z ≤ 4, z3/2 ≤ x ≤ 8}

D′′3 =

{0 ≤ x ≤ 8, x2/3 ≤ z ≤ 4

}=

{0 ≤ z ≤ 4, 0 ≤ x ≤ z3/2

}So

I =

∫∫D′

3

∫ 2

x1/3f(x, y, z) dy dAx,z +

∫∫D′′

3

∫ 2

√zf(x, y, z) dy dAx,z

=

∫ 8

0

∫ x2/3

0

∫ 2

x1/3f(x, y, z) dy dz dx +

∫ 8

0

∫ 4

x2/3

∫ 2

√zf(x, y, z) dy dz dx

=

∫ 4

0

∫ 8

z3/2

∫ 2

x1/3f(x, y, z) dy dx dz +

∫ 4

0

∫ z3/2

0

∫ 2

√zf(x, y, z) dy dx dz

. . . . . .

Outline





Triple Integrals


. . . . . .

Triple Integrals in Cylindrical and Spherical Coordinates ILearning Objectives

◮ To understand and be able to change to cylindrical coordinatesin a triple integral,∫∫∫

E

f(x, y, z) dV

=

∫ β

α

∫ h2(θ)

h1(θ)

∫ u2(r cos θ,r sin θ)

u1(r cos θ,r sin θ)f(r cos θ, r sin θ, z) r dz dr dθ,

where E is the region given by

E = {(x, y, z)|(x, y) ∈ D, u1(x, y) ≤ z ≤ u2(x, y)}

and D = {(r, θ)|α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ)}.

. . . . . .

Triple Integrals in Cylindrical and Spherical Coordinates IILearning Objectives

◮ To understand and be able to change to spherical coordinates ina triple integral,∫∫∫

E

f(x, y, z) dV

=

∫ d

c

∫ β

α

∫ b

af(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ2 sin φ dρ dθ dφ,

where E is a spherical wedge given by

E = {(ρ, θ, φ)|a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d}

. . . . . .

Cylindrical CoordinatesJust add the vertical dimension

.

.x

.y

.z

..

.r.θ

.(r, θ, z)

.z

.x.y

.z

Conversion from cylindrical tocartesian (rectangular):

x = r cos θ y = r sin θ

z = z

Conversion from cartesian tocylindrical:

r =√

x2 + y2

cos θ =xr

sin θ =yr

tan θ =yx

z = z

. . . . . .

Suppose E is a region of (arabic) type 1:

E = { (x, y, z) | (x, y) ∈ D, u1(x, y) ≤ z ≤ u2(x, y) }

where D is a polar region:

D = { (r, θ) | α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ) }

Then∫∫∫E

f(x, y, z) dV =

∫∫D

∫ u2(x,y)

u1(x,y)f(x, y, z) dz dA

=

∫ β

α

∫ h2(θ)

h1(θ)

∫ u2(x,y)

u1(x,y)f(r cos θ, r sin θ, z) dz r dr dθ

. . . . . .

Spherical Coordinateslike the earth, but not exactly

.

.x

.y

.z

..(r, θ, φ)

.

.ρ.φ

.θ.x

.y

.z

Conversion from spherical tocartesian (rectangular):

x = ρ sin φ cos θ

y = ρ sin φ sin θ

z = ρ cos φ

Conversion from cartesian tospherical:

r =√

x2 + y2 ρ =√

x2 + y2 + z2

cos θ =xr

sin θ =yr

tan θ =yx

cos φ =zρ

. . . . . .

The volume element in spherical coordinates

A “spherical box” has area approximately

∆V ≈ (∆ρ)(ρ ∆φ)(ρ sin φ ∆θ)

= ρ2 sin φ∆ρ ∆φ ∆θ

Therefore if E is a spherical region described by

E = { (ρ, θ, φ) | a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d }

we have∫∫∫E

f(x, y, z) dV

=

∫ d

c

∫ β

α

∫ b

af(ρ sin θ cos φ, ρ sin θ sin φ, ρ cos φ)ρ2 sin φ dρ dθ dφ

. . . . . .


Find∫∫∫

Hz3

√x2 + y2 + z2 dV, where H is the solid hemisphere that

lies above the xy-plane and has center the origin and radius 1.

SolutionIn cylindrical coordinates the answer is

∫ 2π

0

∫ 1

0

∫ √1−r2

0z3

√r2 + z2 r dz dr dθ

In spherical coordinates the answer is∫ 2π

0

∫ π/2

0

∫ 1

0(ρ cos φ)3 · ρ · ρ2 sin φ dρ dφ dθ

. . . . . .

Solution (continued)In spherical coordinates the answer is∫ 2π

0

∫ π/2

0

∫ 1

0(ρ cos φ)3 · ρ · ρ2 sin φ dρ dφ dθ

= 2π

∫ 1

0ρ6 dρ ·

∫ π/2

0cos3 φ sin φ dφ

=2π

7

[−1

4cos4 φ

]π/2

0=

π

14

final review ii

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