Calc III Final Solutions

April 2019

1.Points would only be awarded for correct answers. There is no partial credit for any incorrect one.

(a) Find the equation of plane passing through the points A = (0, 1, 3), B = (2, 1,−1), C = (0, 3, 2).Find the distance from the origin to this plane.

Solution. AB = (2, 0,−4), AC = (0, 2,−1). AB×AC = (4, 1, 2). The equation: 4x+ y+ 2z = 7 (3pts).The distance

|0 · 4 + 0 · 1 + 0 · 2− 7||16 + 1 + 4| =

7√21

=

√21

3.

(b) Compute the double integral∫ 1

0(∫√xx

ex/ydy)dx.

Solution.

∫ 1

0

(

∫ √x

x

ex/ydy)dx =

∫ 1

0

∫ y

y2ex/ydxdy =

∫ 1

0

(ye− yey)dy =e

2− 1.

Calculate∫c(x+ y2 + z2)ds, where c is the helix c(t) = (cos t, sin t, t) for 0 ≤ t ≤ 3π.

Solution.

∫

c

(x+ y2 + z2)ds =

∫ 3π

0

(cos t+ sin2 t+ t2)||c′(t)||dt

=

∫ 3π

0

(cos t+ sin2 t+ t2)√

2dt

=√

2(0 +3π

2+ 9π3) =

√2(

3π

2+ 9π3).

Let C be a piecewise smooth closed curve that is the boundary of a surface S, calculate∫C

(2019, 5, 9) ·ds.

Solution. Let F = (2019, 5, 9). By Stokes Theorem,

∫

C

F · ds =

∫∫

S

(O× F ) · dS = 0,

where

O× F =

∣∣∣∣∣∣

i j k∂∂x

∂∂y

∂∂z

2019 5 9

∣∣∣∣∣∣= 0.

1

2 Ca ) Evaluate Sf F . ds,

where F = Csi, y

'

,I ) and S = Lex , y ,z > I x 't y 't 2-2=43

.

Sol I

n = Is Cx, y , 2-7 b/c the surface is the sphere with radius 2 .

Sf F . ds = Sf F - n ds

=

Sf Csi, y

'

, E) . I Cx, y ,

't I IS

=Sf I cast y 't 2-3 ) DS

.

The surface is symmetric ,hence o = Sf odds = Sf yids = Sf Ids

.

As a result,

Sf I cast y 't z'

> ds = I ( Sf odds t Sf yids + Sf Eds ) = o.

So I 2

A parametrization for the surface S is Coutward orientation )

of f ER,

Icf,

f ) = C 2 cos A sin to,

251nF sin &,

2 cost ), o E f Eze

.

Tp = C z cost cost,

2 sin -0 cos f,

-2 sin f)To = ( - 2 sin A sink

,

2cosO-s.lu/0,

o )

Tpx To.

= C 4 sin'

locos -0,4 sin 't sin O,

4 cost sin to )

F C Ect, as ) o Tpi

To

= ( 4050 sin 't,

4 sin'

O sin'

f,

405/0 ) . ( 4 sin'

locos -0,4 sin 't sin O,

4 cost sin to )

= 16 cos'

G sin" to t 16 sin '

f sin"

to t 16 cos'

to sin f .

Hence,

S§ F . ds = So"

) !"

( If cos'

f sin 4ft 16 sin 'f sin

" lot16 cos'

to sinf) dfdf.

After Some c in fact. tons of ) work

,we can get 0 as our answer

.

3) State Stoke’s theorem, and any requirements for it to hold. Use Stoke’s theorem to compute the

flux integrals∫∫S1∇× ~F · d ~S1 and

∫∫S2∇× ~F · d ~S2, where ~F (x, y, z) = (x− y, x+ y, z2 + z3),

S1 = {(x, y, z)|0 ≤ z, z = 9− x2 − y2}, S2 = S1 ∪ {(x, y, z)|z = 0, x2 + y2 ≤ 9}.

a - 5 points) You have to quote Stoke’s theorem as it is found in the textbook. Since there is quite abit to it, we decided to only check for a few key features:

• The vector field ~F is C1 (1 point)

• ∂S is the positive oriented boundary of S (1 point)

•∫∫S∇× ~F · d~S =

∫∂S

~F · d~s (3 points)

b - 10 points) You now have to compute the first integral using Stoke’s theorem. Since this questionwas designed to test your knowledge of Stoke’s theorem, failure to follow this instruction costs 4 points. Analternate rubric is below in the case that Stoke’s theorem was not employed.

It helps to identify the shape in question. S1 is the top of an upside-down parabaloid. One way to seethis is to note that S1 is given by the equation z = 9− r2, so the relationship between z and r is quadratic.

From this picture, we can see that ∂S1 is just the circle of radius 3 that lies in the x−y plane. We can easilyparametrize it as (3 cos(t), 3 sin(t), 0), 0 ≤ t ≤ 2π. Note that that we oriented our circle counterclockwise, inaccordance with how Stoke’s theorem works.1 Applying Stoke’s theorem, we now get

∫

S1

∇× ~F · d ~S1 =

∫

∂S1

~F · d~s =

∫ 2π

0

(3 cos(t)− 3 sin(t), 3 cos(t) + 3 sin(t), 0) · (−3 sin(t), 3 cos(t), 0) dt

=

∫ 2π

0

9 sin2(t) + 9 cos2(t) dt = 18π

6 points were given for correctly working with ∂S1. In particular, 3 points were given for indicating thecorrect region ∂S1, 2 points for a parametrization, and 1 point for saying 0 ≤ t ≤ 2π. 2 points were deductedfor having an incorrect orientation. The remaining 4 points were for evaluating the integral, broken down as2 points for F (c(t)), 1 point for c′(t), and 1 point for dotting them and computing the correct answer.

1Just for your reference, the rule is to you imagine you are a tiny ant walking along ∂S1 in the counterclockwise direction.Also, for you, let ”up” be in the direction S1 is oriented in, the direction of positive flux. Notice that as you walk, the surfaceitself will always be to your left side. Alternatively, if you take even the slightest step to the right, you will fall off of the surface.This tells you that counterclockwise is the correct choice of orientation for ∂S1. Also notice that if you were to instead walkclockwise, then the surface would always be to your right. That is not what you want.

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b - Without Stoke’s, 6 points max) If you went ahead and computed the flux integral, you couldreceive at most 6 of the 10 points for part b. They were distributed as follows:

• Parametrize S1 (1 point)

• Give the correct bounds on your parameters (1 point)

• Compute ∇× ~F = (0, 0, 2) (1 point)

• Compute the normal vector for your parametrization (2 points)

• Compute the correct final answer (1 point)

There were at least three correct ways to parametrize the surface. Some common ways were:

• Φ(r, θ) = (r cos(θ), r sin(θ), 9− r2), 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π

• Φ(z, θ) = (√

9− z cos(θ),√

9− z sin(θ), z), 0 ≤ z ≤ 9, 0 ≤ θ ≤ 2π

• Φ(x, y) = (x, y, 9− x2 − y2), x2 + y2 ≤ 9

Note that in the third parametrization, your coordinates are restricted to lying in a circle of radius 3. Also, becareful of orientation! It turns out that the first and third parametrizations above have the correct outwardorientation, but the second is actually oriented inwards!

c - 5 points) The surface S2 in question is the top of the parabaloid as we had before, but with acircular disk glued to the bottom that closes it off. Indeed, S2 = S1 ∪ {(x, y, z)|z = 0, x2 + y2 ≤ 9}. LetS3 = {(x, y, z)|z = 0, x2 + y2 ≤ 9}, so that S2 = S1 ∪ S3. Looking at the equations, we can see S3 is adisk of radius 3, lying in the x − y plane. So, S1 and S3 snap together perfectly to make a shape with noboundary. Thus, ∂S2 is the empty set. By Stoke’s theorem, the flux integral is 0 .

Alternatively, you could break up S2 into the S1 and S3 and add their flux integrals together. Wecomputed the flux integral of the S1 in part b, it is 18π. Since S3 has the same boundary as S1, it willhave the same flux integral of 18π, right? Well, almost. S2 (by its definition) is oriented outwards. But, ifyou picture the region in your head, you can see that this means S1 will be oriented upwards and S3 willbe oriented downwards! Since S3 is oriented downwards, when you apply the trick in the footnote of theprevious page, you will get a clockwise orientation on ∂S3. This reversed orientation will mean the resultingline integral is −18π. Thus, when we add the two flux integrals, 18π − 18π = 0.

5 points were awarded for the correct answer. 3 points were given if you mistakenly added instead ofsubtracting the two flux integrals and got 18π + 18π = 36π. 1 point was given if you computed the integralover S3 and got 18π as your final answer.

c - Without Stoke’s, 1 point max) If you did not apply Stoke’s theorem in part c, the same penaltyof 4 points would be applied here as well. Thus, at most one point could be received and was given if youobtained the correct answer of 0. However, even if you answered 0, the point was only awarded if you arrivedat it through correct reasoning, mainly by computing the flux integrals over S1 and S3, and noting that theyadd up to 0.

2

4 (1) Evaluate the integral∫C+ x

2019dx+x5y9dy, where C+ is the perimeter of the square [0, 1]×[0, 1] in thecounterclockwise direction. (10 pts) (2) Evaluate the integral

∫cx2019dx+ (y5 + y9)dy, where c : [1, 2]→ R2

is given by c(t) = (et−1, sin(π/t)). (10 pts)

Solution (1) We have∫C+ x

2019dx = 0 since (x2019, 0) is a gradient vector field. On the other hand

∫

C+

x5y9dy =

∫ 1

0

15y9dy −∫ 1

0

05y9dy =

∫ 1

0

y9dy =y10

10|10 = 1/10

(b) Note that (x2019, y5 + y9) is a gradient vector field. In fact, we have

(x2019, y5 + y9) = ∇(x2020/2020 + y6/6 + y10/10)

Set f(x, y) = x2020/2020 + y6/6 + y10/10. By the fundamental theorem of Calculus, we have

∫

c

∇f = f(c(1))− f(c(0))

Note c(0) = (1, 0) and c(1) = (e, 1). So

f(c(1))− f(c(0)) = f(e, 1)− f(1, 0) = (e2020 − 1)/2020 + 4/15

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Problem 5. Let xi ≥ 0, i = 1, 2, . . . , n, be non-negative real numbers such that x1 + x2 + · · · + xn = 1.Let f(x1, . . . , xn) = x1x2 · · ·xn. Use Lagrange Multiplier Strategy to find the maximum value of f when (1)n = 3; (10 pts) (2) n is any fixed positive integer. (10 pts)

(a) Here we want to maximize f(x1, x2, x3) = x1x2x3 subject to the contrainst g(x1, x2, x3) = 1 whereg(x1, x2, x3) = x1 + x2 + x3. First note that since the surface {(x1, x2, x3) ∈ R3 : x1 + x2 + x3 = 1, xi ≥ 0}is a closed and bounded domain, an absolute maximum must occur.

Additionally, notice that on the boundary of the domain, some xi = 0, which makes f = 0. Therefore, nomaximum value occurs at on the boundary. 3 points were taken off for missing this case. Therefore,the maximum values occurs where each xi 6= 0, in which case we employ Langrange Multipliers.

Using Lagrange Multipliers, we know there is a λ ∈ R so that at the maximum we have ∇f = λ∇g.After computing the gradient, this equation becomes:

(x2x3, x1x3, x1x2) = λ(1, 1, 1)

Therefore, λ = x2x3 = x1x3 = x1x2.

Since we are assuming no xi = 0, we my divide by any xi in equations above. So, in particular:

• x2x3 = x1x3 gives x1 = x2

• x1x3 = x1x2 gives x2 = x3

Therefore, x1 = x2 = x3. From the contraint g = 1 we have 3x1 = 1, thus 13 = x1 = x2 = x3. Thus

f(x1, x2, x3) = ( 13 )3 = 1

27 which must be the maximum.

(b) In the more general case, we want to maximize f(x1, x2, ..., xn) = x1x2 · · ·xn subject to the contrainstg(x1, x2, ..., xn) = 1 where g(x1, x2, ..., xn) =

∑xi. First note that since the surface {(x1, x2, x3) ∈ R3 :

x1 + x2 + ...xn = 1, xi ≥ 0} is a closed and bounded domain, an absolute maximum must occur.

Additionally, notice that on the boundary of the domain, some xi = 0, which makes f = 0. Therefore, nomaximum value occurs at on the boundary. 2 points were taken off for missing this case. Therefore,the maximum values occurs where each xi 6= 0, in which case we employ Langrange Multipliers.

Using Lagrange Multipliers, we know there is a λ ∈ R so that at the maximum we have ∇f = λ∇g.After computing the gradient, this equation becomes:

(f

x1,f

x2, ...,

f

xn) = λ(1, 1, , ..., 1)

Therefore, λ = fx1

= fx2

= ... = fxn

. Note these are well-defined fractions since we assuming each xi 6= 0.

7

This also implies f 6= 0, so we may divide by f in each equation to get:

1

x1=

1

x2= ... =

1

xn

Thus, x1 = x2 = ... = xn.. From the contraint g = 1 we have nx1 =∑

1≤i≤n x1 = 1, thus 1n = x1 = x2 =

...xn. Thus f(x1, x2, ..., xn) = ( 1n )n which must be the maximum.

8