ExamI - NYU Courantcem9/old_teaching/su16_3140/exam1.pdf · 2018-11-24 · ExamI...

Exam IMath 3140 - Vector Calc & PDEs

June 17, 2016

Answer each question completely in the area below. Show all work and explain your reasoning. Ifthe work is at all ambiguous, it is considered incorrect. No phones, calculators, or notes are

allowed. Anyone found violating these rules or caught cheating will be asked to leaveimmediately. Point values are in the square to the left of the question. If there are any other

issues, please ask the instructor.

By signing below, you are acknowledging that you have read and agree to the above paragraph, aswell as agree to abide University Honor Code:

Name:

Signature:

uID:

SolutionsQuestion Points Score

1 15

2 15

3 5

4 15

5 20

6 20

7 15

Total: 105

Note: There are 7 questions on the exam with 105 points available but the exam will be graded outof 100.

Math 3140: Exam I June 17, 2016

Potentially Useful Information

Cylindrical Coordinates

x = r cos θ, y = r sin θ, z = z, r2 = x2 + y2

Spherical Coordinates

x = ρ sinφ cos θ, y = ρ sinφ sin θ, z = ρ cosφ, ρ2 = x2 + y2 + z2

Surfaces

n =ru × rv‖ru × rv‖

, dS = ‖ru × rv‖dA

Useful cryptic hint when r(x, y) = 〈x, y , g(x, y)〉

rx × ry = 〈−gx ,−gy , 1〉.

Green’s Theorem∮C

F · dr =∫∫

D

(∇× F) · k dA∮C

F · n ds =∫∫

D

∇ · F dA

Stokes’ Theorem∮C

F · dr =∫∫

S

(∇× F) · dS

Divergence Theorem∫∫S

F · dS =∫∫∫

E

∇ · F dV

2/12 /0 pts


1.15 Ethan, a dear friend of yours has landed himself in jail and calls you begging for $250 to bail himout. You and another friend, Greg, hatch a plan to come up with the money: go to a casino inWendover.

You are a solid gambler, but get nervous when you start making a lot of money, so the probabilitydensity function of your winnings (denoted X) is

f1(x) =

{15000x 0 ≤ x ≤ 1000 otherwise.

Greg is a way more experienced gambler and is definitely going to make more. However, once hemakes $200, the casino assumes he is cheating and kicks him out with nothing, so the probabilitydensity function of Greg’s winnings (denoted Y ) is

f2(y) =

{1

200(e−1)ey/200 0 ≤ y ≤ 200

0 otherwise.

What’s the probability that you and Greg earn enough (combined) to bail Ethan out of jail?

Set up, but do not compute the integral that describes the answer.

Solution: This is modified from a homework problem and was explicitly mentioned to beon the exam.

It’s clear from the setup of the problem that the two random variables (your earnings and Greg’searnings) are independent, meaning that their joint probability density f (x, y) is just the productof the two probability densities. That is:

f (x, y) = f1(x)f2(y) =

{1

10000(e−1)xey/200 0 ≤ x ≤ 100, 0 ≤ y ≤ 200

0 otherwise.

Thus, to compute the probability that a set of outcomes occurs, we describe this set of outcomesas a region D in 2D space and then

probability outcome is in D =∫∫

D

f (x, y) dA.

We now just need to figure out the region D the problem is describing. The first piece ofinformation is that you and Greg must make enough money to bail Ethan out. Symbolically, thiscorresponds to:

X + Y ≥ 250.

That is, combined, you must make more than $250. We also have two other constraints basedon how much each of you can earn:

X ≤ 100, Y ≤ 200.

Note, these three inequalities define a region D in the XY plane:

3/12 /15 pts


X = 100

D

X + Y = 250

Y = 200

20 40 60 80 100 120 140

120

140

160

180

200

220

240

We could set this up as either type integral. Let’s for instance, take vertical slices. We see thelower limit of y is always y = 250 − x and the upper limit is y = 200. However, the boundsof x are a little tricky. The upper bound is x = 100, but the lower bound is the intersection ofx + y = 250 and y = 200 which occurs at x = 50, thus we have:∫∫

D

f (x, y) dA =

∫ 10050

∫ 200250−x

1

10000(e − 1)xey/200 dy dx.

If you are curious, the answer ends up being approximately 15%, so Ethan has a reasonable shotof making it out of the slammer!

4/12 /0 pts


2.15 Set up, but do not compute the following integral in spherical coordinates:

∫∫∫V

f (x, y , z) dV,

where f (x, y , z) = xy , and V is the region below the spherex2 + y2 + z2 = z and above the cone z =

√x2 + y2.

886 CHAPTER 12 MULTIPLE INTEGRALS

Formula 4 says that we convert a triple integral from rectangular coordinates to spher-ical coordinates by writing

using the appropriate limits of integration, and replacing by . This isillustrated in Figure 8.

This formula can be extended to include more general spherical regions such as

In this case the formula is the same as in (4) except that the limits of integration for areand .

Usually, spherical coordinates are used in triple integrals when surfaces such as conesand spheres form the boundary of the region of integration.

Evaluate where is the unit ball:

SOLUTION Since the boundary of is a sphere, we use spherical coordinates:

In addition, spherical coordinates are appropriate because

Thus (4) gives

Note: It would have been extremely awkward to evaluate the integral in Example 3 with-out spherical coordinates. In rectangular coordinates the iterated integral would have been

A volume that is easier in spherical coordinates Use spherical coordinatesto find the volume of the solid that lies above the cone and below thesphere . (See Figure 9.)

SOLUTION Notice that the sphere passes through the origin and has center . Wewrite the equation of the sphere in spherical coordinates as

The equation of the cone can be written as

! cos " ! s!2 sin 2" cos 2# $ !2 sin2" sin2# ! ! sin "

! ! cos "or!2 ! ! cos "

(0, 0, 12 )x 2 $ y 2 $ z2 ! z

z ! sx 2 $ y 2

EXAMPLE 4v

e !x2$y2$z2"3#2

dz dy dxy1

%1ys1%x 2

%s1%x 2 ys1%x2%y2

%s1%x2%y2

! [%cos "]0& !2&" [ 1

3e!3 ]0

1! 4

3& !e % 1"

! y&

0 sin " d" y2&

0d# y1

0!2e !3 d!

yyyB

e !x2$y2$z2"3#2

dV ! y&

0y2&

0y1

0e!! 2"3#2

!2 sin " d! d# d"

x 2 $ y 2 $ z2 ! !2

B ! $!!, #, "" % 0 ' ! ' 1, 0 ' # ' 2&, 0 ' " ' &&

B

B ! $!x, y, z" % x 2 $ y 2 $ z2 ' 1&BxxxB e !x2$y2$z2"3#2

dV,EXAMPLE 3v

t2!#, ""t1!#, ""!

E ! $!!, #, "" % ( ' # ' ), c ' " ' d, t1!#, "" ' ! ' t2!#, ""&

!2 sin " d! d# d"dV

z ! ! cos "y ! ! sin " sin #x ! ! sin " cos #

FIGURE 8Volume element in sphericalcoordinates: dV=∏ @ sin ˙ d∏ d¨ d˙

z

0

xyd¨

˙

∏ sin ˙ d¨

∏

∏ d˙

d∏

FIGURE 9

(0, 0, 1) ≈+¥+z@=z

z=œ„„„„„ ≈+¥

y x

z

Solution: This exact problem was done in class.

To begin, we first convert the two key surfaces (the sphere, the cone) into spherical coordinatesusing the conversions provided on the front page.

For the sphere, we have:

x2 + y2 + z2 = z ⇔ ρ2 = ρ cosφ =⇒ ρ = cosφ.

For the cone:

z =√x2 + y2 ⇔ ρ cosφ =

√ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ = ρ sinφ

√sin2 θ + cos2 θ,

which simplifies toρ cosφ = ρ sinφ.

When is this true? Only when φ = π/4. Thus, the equation of the cone is φ = π/4, which isexactly the form of a cone in spherical coordinates, which is assuring.

We can now construct our triple integral. Note that ρ varies from the inside of the cone to theoutside of the sphere, meaning that its bounds are ρ = 0 and ρ = cosφ. The angle φ variesfrom vertical, φ = 0, to the outside of the cone, φ = π/4. We also have a full rotation around,meaning the range of θ is θ = 0 to θ = 2π.

We also need to convert f (x, y) = xy into spherical coordinates, which becomes:

f (ρ sinφ cos θ, ρ sinφ sin θ) = ρ2 sin2 φ cos θ sin θ.

Lastly, we recall the volume element in spherical:

dV = ρ2 sinφdρ dφ dθ.

Piecing all this information, our triple integral becomes:∫∫∫V

f (x, y , z) dV =

∫ 2π0

∫ π4

0

∫ cosφ0

(ρ2 sin2 φ cos θ sin θ

)ρ2 sinφ dρ dφ dθ.

5/12 /15 pts


3.5 The figure shows a vector field F and two curves C1 and C2. Are the line integrals∮C1F · dr and∮

C2F · dr positive, negative, or zero? Explain.

SECTION 13.2 LINE INTEGRALS 923

17. Let be the vector field shown in the figure.(a) If is the vertical line segment from to

, determine whether is positive, negative,or zero.

(b) If is the counterclockwise-oriented circle with radius 3and center the origin, determine whether is pos-itive, negative, or zero.

18. The figure shows a vector field and two curves and .Are the line integrals of over and positive, negative, or zero? Explain.

19–22 Evaluate the line integral , where is given bythe vector function .

19. ,,

20. ,,

21. ,,

22. ,,

23–26 Use a calculator or CAS to evaluate the line integralcorrect to four decimal places.

23. , where and , 1 ! t ! 2r!t" ! e t i " e#t2

jF!x, y" ! xy i " sin y jxC F ! dr

0 ! t ! $r!t" ! t i " sin t j " cos t kF!x, y, z" ! z i " y j # x k

0 ! t ! 1r!t" ! t 3 i # t 2 j " t kF!x, y, z" ! sin x i " cos y j " xz k

0 ! t ! 1r!t" ! t 2 i " t 3 j " t 2 kF!x, y, z" ! !x " y" i " !y # z" j " z2 k

0 ! t ! 1r!t" ! 11t 4 i " t 3 jF!x, y" ! xy i " 3y 2 j

r!t"CxC F ! dr

y

x

C¡C™

C2C1FC2C1F

y

x0 1

1

2 3

2

3

_3 _2 _1

_3

_2

_1

xC2 F ! dr

C2

xC1 F ! dr!#3, 3"

!#3, #3"C1

F 24. , where and ,

25. , where has parametric equations ,, ,

26. , where has parametric equations , ,,

27–28 Use a graph of the vector field F and the curve C to guesswhether the line integral of F over C is positive, negative, or zero.Then evaluate the line integral.

27. ,is the arc of the circle traversed counter-

clockwise from (2, 0) to

28. ,

is the parabola from to (1, 2)

29. (a) Evaluate the line integral , whereand is given by

, .; (b) Illustrate part (a) by using a graphing calculator or com-

puter to graph and the vectors from the vector field corresponding to , , and 1 (as in Figure 13).

30. (a) Evaluate the line integral , whereand is given by

, .; (b) Illustrate part (a) by using a computer to graph and

the vectors from the vector field corresponding to and (as in Figure 13).

31. Find the exact value of , where is the curve withparametric equations , , ,

.

32. (a) Find the work done by the force fieldon a particle that moves once

around the circle oriented in the counter-clockwise direction.

(b) Use a computer algebra system to graph the force fieldand circle on the same screen. Use the graph to explainyour answer to part (a).

33. A thin wire is bent into the shape of a semicircle, . If the linear density is a constant , find

the mass and center of mass of the wire.

34. A thin wire has the shape of the first-quadrant part of the circle with center the origin and radius . If the density function is , find the mass and center of mass of the wire.

35. (a) Write the formulas similar to Equations 4 for the centerof mass of a thin wire in the shape of a spacecurve if the wire has density function .%!x, y, z"C

!x, y, z "

%!x, y" ! kxya

kx & 0x 2 " y 2 ! 4

CAS

x 2 " y 2 ! 4F!x, y" ! x 2 i " xy j

0 ! t ! 2$z ! e#ty ! e#t sin 4 tx ! e#t cos 4 t

CxC x 3y 2z dsCAS

' 12t ! '1

C#1 ! t ! 1r!t" ! 2t i " 3t j # t 2 k

CF!x, y, z" ! x i # z j " y kxC F ! dr

1#s2 t ! 0C

0 ! t ! 1r!t" ! t 2 i " t 3 jCF!x, y" ! e x#1 i " xy j

xC F ! dr

!#1, 2"y ! 1 " x 2C

F!x, y" !x

sx 2 " y 2 i "

y

sx 2 " y 2 j

!0, #2"x 2 " y 2 ! 4C

F!x, y" ! !x # y" i " xy j

CAS

0 ! t ! 1z ! e#ty ! t 2x ! tCxC ze#xy ds

0 ! t ! 5z ! t 4y ! t 3x ! t 2CxC x sin!y " z" ds

0 ! t ! $r!t" ! cos t i " sin t j " sin 5t kF!x, y, z" ! y sin z i " z sin x j " x sin y kxC F ! dr

Solution: The angle between F and C1 is always between zero and 90◦, so the work done alongC1 is positive.

The second integral is a little trickier, but we notice that the work done along (roughly) the firsthalf of C2 is positive, and along the secondhalf, negative. However, since the field is strongeralong the second half, the overall work done will be negative. If you wrote anything vaguelycoherent for the second one, you probably received credit.

6/12 /5 pts


4. True or false? Either way, explain your reasoning.

(a)5 div curlF = 0 for all vector fields F.

(b)5 If L is the line segment connecting (−1, 0) and (1, 0), C+ is the upper half of the unit circleand C− is the lower half, and F satisfies∫

L

F · dr =∫C+F · dr =

∫C−F · dr = 0,

then F must be conservative.

(c)5 grad divF = 0 all vector fields F.

Solution: This question was taken directly from a practice exam.

(a) True. We discussed this in class. One way to see it is if we think symbolically: ∇· (∇×F),we know that × produces an orthogonal vector to both ∇ and F, and therefore the dotproduct of ∇ with this will be 0.

(b) False. The integral being equal on three paths does not mean it is equal on all paths, whichis the requirement for F to be conservative.

(c) False. No reason to believe this is true. For instance, consider F = 〈x2, 0, 0〉, then∇ · F = 2x and then ∇(∇ · F) = 2 6= 0.

7/12 /15 pts


5.20 Use Green’s Theorem to evaluate the line integral:∮C

xy2 dx + 2x2y dy ,

where C is the positively oriented triangle with vertices (0, 0), (2, 2), (2, 4).

Solution: Green’s Theorem (ignoring all the technical conditions), says that if F = 〈P,Q〉, then∮C

F · dr =∮C

P dx +Q dy =

∫∫D

[∂Q

∂x−∂P

∂y

]dA.

Here, we see that P = xy2 and Q = 2x2y meaning that Qx = 4xy and Py = 2xy and ourintegrand is Qx − Py = 4xy − 2xy = 2xy .

−2 −1 1 2 3 4

2

4

y = x

y = 2xx = 2

x

y

To know the bounds of integration, we must draw the region. We could set this up as eithertype 1 or type 2, but I think type 1 is easier here. Note that x varies between 0 and 2 and theny is bounded by the two sides of the triangle described by y = 2x and y = x , thus:∫∫

D

[∂Q

∂x−∂P

∂y

]dA =

∫ 20

∫ 2xx

2xy dy dx

=

∫ 20

[xy2]y=2xy=x

dx

=

∫ 20

3x3 dx

=

[3

4x4]x=2x=0

= 12.

8/12 /20 pts


6.20 Using Stokes’ theorem, fully set up but do not evaluate an integral equivalent to∮C

F · dr where F = 〈xy , 2z, 3y〉,

and C is the intersection of the plane x + z = 5 and the cylinder x2 + y2 = 9.

Solution: This question was taken directly from the homework and is very similar to aproblem done in class.

Stokes’ theorem says that instead of computing the line integral over C, we can instead computethe surface integral of S for any surface S with C as a boundary. Here, the natural surface toconsider is the cylinder x2 + y2 = 9 with top sliced by the plane x + z = 5.

Citing Stokes’ theorem: ∮C

F · dr =∫∫

S

∇× F) dS.

Thus, we need to compute the curl of F :

∇× F =

∣∣∣∣∣∣i j k

∂x ∂y ∂zxy 2z 3y

∣∣∣∣∣∣ = 〈1, 0,−x〉.We need now to parameterize our surface. This (like basically every problem we did) has a nicenatural parameterization: r(x, y) = 〈x, y , g(x, y)〉 where g(x, y) is the top of our cylinder, whichclearly is the plane z = 5− x and the domain D of x, y is the circle x2 + y2 ≤ 9.We now recall to actually compute this surface integral, we recall (or see the equation sheet)∫∫

S

F · S =∫∫

D

F · n dS =∫∫

D

F ·(ru × rv)‖ru × rv‖

‖ru × rv‖ dA =∫∫

D

F · (ru × rv) dA.

Thus, with all surface integrals, it just boils down to computing ru× rv . Here, we could computethis, or use the hint on the formula sheet to find

rx × ry = 〈−gx ,−gy , 1〉 = 〈1, 0, 1〉,

recalling that g(x, y) = 5− x . Thus, our integral becomes∫∫S

∇× F · dS =∫∫

D

(∇× F) · (rx × ry ) dA =∫∫

D

〈1, 0− x〉 · 〈1, 0, 1〉 dA =∫∫

D

(1− x) dA.

Here, since D is the circle of radius 3, we convert this to polar to get our final answer∫∫D

(1− x) dA =∫ 2π0

∫ 30

(1− r cos θ) r dr dθ.

9/12 /20 pts


7.15 Compute ∫∫S

F · dS, where F = 〈5x + 6xy , xz − 3y2, cos xy2〉,

and S is the surface of a solid cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1, orientedpositively.

Hint: you probably want to use a theorem for this one!

Solution: This question was taken directly from a practice exam with some slight numberschanged. The divergence theorem says that∫∫

S

F · dS =∫∫∫

E

∇ · F dV.

Here, we have that∇ · F = 5 + 6y − 6y + 0 = 5,

Thus, ∫∫S

F · dS =∫∫∫

E

∇ · F dV =∫∫∫

E

5 dV = 5

∫∫∫E

dV = 5V (E),

since integrating∫∫∫

E 1 dV just is the volume of E. In this case, E is just the box described inthe problem, so V (E) = 13 = 1, so∫∫

S

F · dS =∫∫∫

E

∇ · F dV = 5(13) = 5.

10/12 /15 pts


Bonus Questions8. (Politics) While touring Estonia in 2004 together, Hillary was reportedly challenged by the 2008

presidential nominee (known for his lack of arm mobility) to a drinking contest. She won withfour shots of vodka. Who was the challenger?

Solution: John Mccain.

http://www.telegraph.co.uk/news/2013729/US-elections-How-Hillary-Clinton-beat-John-McCain-at-vodka-drinking.html

9. (Other) When Gus Levy took over as senior partner of Goldman Sachs in 1969, he coined aphrase which has been said to be Goldman’s unofficial philosophy. Which is the phrase?

A. Kill or be killed.

B. Long term greedy.

C. The customer is always right . . . to his face.

D. Profit before family, before country, before God.

Solution: B. Long term greedy.https://www.washingtonpost.com/blogs/ezra-klein/post/at-goldman-short-term-greed-vs-long-term-greed/2011/08/25/gIQAxFhhES_blog.html

10. (Math) Are either of the vector fields F shown below conservative? Why or why not?

SECTION 13.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS 933

16. ,: , , ,

17. ,: ,

18. ,: ,

19–20 Show that the line integral is independent of path and eval-uate the integral.

19. ,is any path from to

20. ,is any path from to

21. Suppose you’re asked to determine the curve that requires theleast work for a force field to move a particle from onepoint to another point. You decide to check first whether isconservative, and indeed it turns out that it is. How wouldyou reply to the request?

22. Suppose an experiment determines that the amount of workrequired for a force field to move a particle from the point

to the point along a curve is 1.2 J and thework done by in moving the particle along another curve

between the same two points is 1.4 J. What can you sayabout ? Why?

23–24 Find the work done by the force field in moving anobject from to .

23. ; ,

24. ; ,

25–26 Is the vector field shown in the figure conservative?Explain.

a. b.

27. If , use a plot to guesswhether is conservative. Then determine whether yourguess is correct.

FF!x, y" ! sin y i ! !1 ! x cos y" jCAS

y

x

y

x

Q!2, 0"P!0, 1"F!x, y" ! e"y i " xe"y j

Q!2, 4"P!1, 1"F!x, y" ! 2y 3#2 i ! 3xsy j

QPF

FC2

FC1!5, "3"!1, 2"

F

FF

!1, 2"!0, 1"CxC !1 " ye"x " dx ! e"x dy

!2, ##4"!1, 0"CxC tan y dx ! x sec2 y dy

0 $ t $ 1r!t" ! t i ! t 2 j ! t 3 kCF!x, y, z" ! e y i ! xe y j ! !z ! 1"ez k

0 $ t $ #r!t" ! t 2 i ! sin t j ! t kCF!x, y, z" ! y 2 cos z i ! 2xy cos z j " xy 2 sin z k

0 $ t $ 1z ! 2t " 1y ! t ! 1x ! t 2CF!x, y, z" ! !2xz ! y2" i ! 2xy j ! !x 2 ! 3z2" k 28. Let , where . Find curves

and that are not closed and satisfy the equation.

(a) (b)

29. Show that if the vector field is conser-vative and , , have continuous first-order partial deriva-tives, then

30. Use Exercise 29 to show that the line integralis not independent of path.

31–34 Determine whether or not the given set is (a) open, (b) connected, and (c) simply-connected.

31. 32.

33.

34.

35. Let .

(a) Show that .(b) Show that is not independent of path.

[Hint: Compute and , where and are the upper and lower halves of the circle

from to .] Does this contradictTheorem 6?

36. (a) Suppose that is an inverse square force field, that is,

for some constant , where . Find thework done by in moving an object from a point along a path to a point in terms of the distances and

from these points to the origin.(b) An example of an inverse square field is the gravita-

tional field discussed in Example 4in Section 13.1. Use part (a) to find the work done by the gravitational field when the earth moves from aphelion (at a maximum distance of km from the sun) to perihelion (at a minimum distance of

km). (Use the values kg,kg, and

(c) Another example of an inverse square field is the electricforce field discussed in Example 5 inSection 13.1. Suppose that an electron with a charge of

C is located at the origin. A positive unitcharge is positioned a distance m from the electronand moves to a position half that distance from the elec-tron. Use part (a) to find the work done by the electricforce field. (Use the value .)% ! 8.985 & 10 9

10"12"1.6 & 10"19

F ! %qQr#$ r $3

N'm2#kg2."G ! 6.67 & 10"11M ! 1.99 & 1030m ! 5.97 & 10241.47 & 108

1.52 & 108

F ! "!mMG "r#$ r $3

d2

d1P2

P1Fr ! x i ! y j ! z kc

F!r" !cr

$ r $3

F

!"1, 0"!1, 0"x 2 ! y 2 ! 1C2

C1x C2

F ! drx C1

F ! drxC F ! dr(P#(y ! (Q#(x

F!x, y" !"y i ! x j

x 2 ! y 2

%!x, y" $ !x, y" " !2, 3"&

%!x, y" $ 1 $ x 2 ! y 2 $ 4, y ) 0&

%!x, y" $ 1 * $ x $ * 2&%!x, y" $ 0 * y * 3&

xC y dx ! x dy ! xyz dz

(Q(z

!(R(y

(P(z

!(R(x

(P(y

!(Q(x

RQPF ! P i ! Q j ! R k

yC2

F ! dr ! 1yC1

F ! dr ! 0

C2

C1f !x, y" ! sin!x " 2y"F ! + f

11/12 /0 pts


Solution: The key to this problem is to consider a closed loop integral in each of these. Weknow that if F is conservative, any closed loop integral must be 0. If we consider the first vectorfield, a closed loop integral (say around the origin) definitely be negative since the arrows alwayspoint against the curve.

The second one, although a little ambigious, does have this feature. The arrows in some sensecancel each other out for any closed loop.

12/12 /0 pts

ExamI - NYU Courantcem9/old_teaching/su16_3140/exam1.pdf · 2018-11-24 · ExamI...

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