MATH 52 FINAL EXAM JUNE 5, 2009
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MOST PROBLEMS ARE WORTH 10 POINTS. PROBLEM TEN IS 12 POINTS AND
PROBLEM ELEVEN IS 8 POINTS. THE TOTAL IS 120 POINTS.
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SIGNATURE:
Prob/[pts]
Score
NAME:
Prob/[pts]
Score
Prob/[pts]
1 [10]
5 [10]
9 [10]
2 [10
6 [10]
10 [12]
3 [10]
7 [10]
11 [8]
4 [10]
8 [10]
12 [10]
1
Score
1. The sum of the two integrals
Z 0 Z 2
Z
2
7
3x cos(y )dydx +
√
−4
−x
0
4
Z
2
√
3x2 cos(y 7 )dydx
x
represents the integral of 3x2 cos(y 7 ) over a region D in the xy-plane.
(a)[4]. Sketch the region D.
The region is ‘trumpet’ shaped, with the top is the segment {(x, 2) | − 4 ≤ x ≤ 4}, the
vertex is the origin, and the sides are given by x = −y 2 and x = y 2
(b)[4]. Reverse the order of integration and write the integral over D as one double integral
R d R b(y)
[Check that you have sketched D correctly.]
in the form c a(y) 3x2 cos(y 7 ) dxdy.
R 2 R y2
0
−y 2
3x2 cos(y 7 ) dxdy
(c)[2]. Evaluate the integral in (b) or the two integrals above part (a), your choice.
R2
R2
2
Above = 0 x3 cos(y 7 )|y−y2 dy = 0 2y 6 cos(y 7 ) dy = 72 sin(y 7 )|20 = 72 sin(27 )
2
2(a)[5]. Let W be the solid object in the half space x ≥ 0 bounded by paraboloids z =
27 − 2x2 − 2y 2 and z = x2 + y 2 and by the yz-plane. Assume the density of the object
is given by δ(x, y, z) = Rx.R RSet up but do not evaluate a triple integral in rectangular
coordinates in the order
( ) dzdydx that gives the mass of the object.
R 3 R √9−x2
0
√
− 9−x2
R 27−2x2 −2y2
x2 +y 2
x dzdydx
(b)[5]. Set up but do not evaluate an integral in cylindrical coordinates r, θ, z that gives
the mass of the same object.
R π/2 R 3 R 27−2r2
−π/2
0
r2
rcos(θ) r dzdrdθ
3
2
2
3. Let C be the ellipse x16 + y9 = 1 in the xy-plane. Let F~ be the vector field given by
F~ = (3y, x) = 3y ~i + x ~j.
H
(a)[3]. Compute the flux C F~ · ~n ds where ~n is the outward unit normal to curve C in the
xy-plane. [Hint: Does the divergence theorem apply?]
~
div(
R R F ) = 0+0 = 0 at all points of the plane, so the divergence theorem applies.
R 0 dA = 0, where R is the region bounded by the ellipse.
H
C
F~ ·~nds =
H
(b)[3]. Compute the circulation C F~ · ~t ds, where ~t is the unit tangent vector to curve C
oriented clockwise. [Hint: Green’s theorem.]
Write F~ = (P, Q) = (3y, x). We will use Green’s theorem, but C is oriented clockwise, so
we must change the usual sign.
H
RR
RR
~ ~
c F · ds = −
R Qx − Py dA = −
R −2 dA = 24π, since the ellipse with major and
minor axes of length 3 and 4 has area 12π.
(c)[4]. Using the parametrization (4cos(θ), 3sin(θ)), 0 ≤ θ ≤ 2π, for the ellipse C, write
down but do not evaluate an explicit integral that gives the area of one side of a fence whose
base is curve C and whose height above the xy-plane is given by the formula 10 − y.
The fence area is the scalar integral of the height function over the curve.
p
R 2π
R 2π
2
2
~
0 (10 − 3sin(θ))||T |kdθ = 0 (10 − 3sin(θ)) 16sin (θ) + 9cos (θ) dθ, since the tangent
vector of the curve (4cos(θ), 3sin(θ)) is T~ = (−4sin(θ), 3cos(θ)).
4
4(a)]5]. Consider the change of variables u = xy, v = y/x. Find the Jacobian determinant
∂(x, y)
∂(u, v)
as a function of u and v. [Hint: First find the reciprocal as a function of x and y.]
First,
∂(u,v)
∂(x,y)
= det
y
x
2
−y/x 1/x
= 2(y/x) = 2v.
∂(x,y)
So ∂(u,v)
= 1/2v, by properties of determinants and the rules for derivatives of an inverse
function.
(b)[5]. Let R be the domain in the first quadrant of the xy-plane bounded by the hyperbolas
xy = 1 and xy = 3 and
R R by the lines y = 2x and y = 5x. Use the change of variables in
part (a) to compute
R y/x dxdy. [Hint: R corresponds to a very simple region in the
uv-plane]
xy = 1, xy = 3 is the same as u = 1, u = 3. Also, y = 2x, y = 5x is the same as
v = 2, v = 5.
We are integrating the function y/x = v, which we do by change of variables over to the
uv-plane.
RR
R3 R5
R3R5
R3R5
∂(x,y)
1
v
y/x
dxdy
=
v
1/2v
dvdu
=
dvdu
=
2
1
R
1 2 1/2 dvdu = 2 2 3 =
1 2
∂(u,v)
3.
5
5(a)[4]. The vector field F~ = (3x2 y + z 2 + y, x3 + 2yz + x, 2zx + y 2 + 1) is conservative.
Find a potential function f (x, y, z) for F~ .
First get ∂f /∂x right, then add terms in y and z to get ∂f /∂y right, and finally add terms
in z to get ∂f /∂z right.
f (x, y, z) = (x3 y + z 2 x + yx) + (y 2 z) + z.
(b)[3]. If C is the curve (t, t2 , t3 ), 0 ≤ t ≤ 1, compute the vector line integral
where F~ is the vector field of part (a). [Use any method.]
R
C
~
F~ · ds,
~ = f (1, 1, 1) − f (0, 0, 0) = 5 − 0 = 5, by path independence of line integrals of
∇f
conservative ( = gradient) vector fields.
R
C
[f (x, y, z) is the potential function from part (a).]
(c)[3]. If the vector field F~ in part (a) represents a force field, determine the work done
by F~ on a particle that moves counterclockwise once around the unit circle x2 + y 2 = 1.
Explain.
The work is the line integral
H
C
~ = 0, since F~ = ∇f
~ is a conservative vector field.
F~ ·ds
[Here C is the closed unit circle.]
6
6. Let S be the portion of the sphere x2 + y 2 + z 2 = 25 that lies above the plane z = −3.
[So S consists of more than half of a sphere.] Let F~ = (x+3y 2 +z 2 , x2 +y−z 2 , xy−2z).
(a)[2]. Calculate div(F~ ). At what points of R3 does your calculation hold?
div(F~ ) = ∂P/∂x+∂Q/∂y +∂R/∂z = 1+1−2 = 0. The function and its partial derivatives
are defined on all of R3 , so div(F~ ) = 0 at all points of R3 .
RR
~ n dS, where the unit normal is given by ~n = 1 (x, y, z)
(b)[8]. Compute the flux
5
S F ·~
at points of S. Explain your work very clearly. For your own protection, NO credit will be
given for attempts that try to parametrize S.
[Hint: S is not a closed surface. But S together with a disk in some plane does form a
closed surface. You can also use symmetry and knowledge of certain areas to evaluate
certain integrals without actually integrating.]
We first replace S by a closed surface S ∪ D, where D is the disk x2 + y 2 + z 2 ≤ 16, z = −3.
Since div(F~ ) = 0 on and inside this closed surface, we get
Z Z
Z Z
Z Z
~
~
~
~
~
0=
F · dS =
F · dS +
F~ · dS,
S∪D
S
D
where we use the outward normal of the closed surface on both S and D.
~
~
~
On D the outward
− 2z), and after a few sign
R R normal is −kR=R (0, 0, −1), so F · −k =R −(xy
R
~
~
changes we get
S F · dS = −
D −(xy − 2z) dA =
D xy + 6 dA, since z = −3 on
D.
RR
RR
By symmetry,
D xy dA = 0. Thus the answer is
D 6 dA = 6(Area D) = 6(16π) =
96π.
7
~
7. Consider the conical surface S parametrized as X(u,
v) = (usin(v), u, ucos(v)) with
0 ≤ u ≤ 4, 0 ≤ v ≤ 2π. The vertex of S is at the origin, and the axis of the cone S is
along the positive y-axis.
~ to surface S produced by this parametriza(a)[5]. Consider the standard normal vector N
~ and ||N
~ ||, in terms of u and v. Does N
~ point toward or away from the
tion. Compute N
~
axis of S? [N is not a unit normal. The order of the parameter variables is (u, v).]
~i
~

N = det
sin(v)
ucos(v)


~k
~j
1
cos(v)  = (−usin(v), u, −ucos(v))
0 −usin(v)
~ points from the cone toward the ySince the y coordinate u is positive, the normal N
axis.
√
p
√
~ || = u2 sin2 (v) + u2 + u2 cos2 (v) = 2u2 = 2u.
||N
√
(b)[5]. The surface area of S is 16 2π. Consider the centroid (x̄, ȳ, z̄) of S. [Centroid
means center of mass when density δ = 1. ] Two of the three centroid coordinates are
obvious. Which two, and what are their values? Explain how you would calculate the third
centroid coordinate in terms of an explicit integral in terms of u and v. You do not need
to evaluate.
Since the object is symmetrical about the y-axis and since the density is constant, the
center of mass lies on the y-axis. This means x̄ = z̄ = 0.
To calculate ȳ, we need to divide the moment Rabout
the plane y = 0 by the surface area
R
of the cone. The moment is the scalar integral
y
dS. So the answer is
S
1
√
16 2π
Z
0
2π
Z
4
~ || dudv =
u||N
0
8
1
√
16 2π
Z
0
2π
Z
0
4
√
u( 2u) dudv
8. In this problem, we consider the plane vector field F~ = (P (x, y), Q(x.y)) = ( x2−y
, x ),
+y 2 x2 +y 2
which satisfies ∂Q/∂x = ∂P/∂y at all points where F~ is defined. Parts (b) and (c) of Problem 8 are on the next page.
H
~
(a)[4]. If C is the unit circle x2 + y 2 = 1 oriented counterclockwise, calculate C F~ · ds.
H
~ = 0?
Why doesn’t Green’s theorem give the result C F~ · ds
We parametrize the circle counterclockwise as (cos(t), sin(t), 0 ≤ t ≤ 2π, with tangent vector (−sin(t), cos(t)). On C, the vector field is F~ = (P, Q) = (−sin(t), cos(t)). Then
I
C
~ =
F~ · ds
Z
2π
Z
(−sin(t), cos(t)) · (−sin(t), cos(t))dt =
0
2π
1dt = 2π
0
RR
The Green’s formula
R (Qx −Py ) dA for the circulation integral does not apply since the
partial derivatives Qx , Py are not continuous in the region R bounded by circle C.
9
8(b)[3]. Sketch the closed oriented polygonal curve C 0 with vertices
(3, 0), (3, −3), (0, −3), (−3, 0), (0, 3), (3, 0)
in that order. Explain
how Green’s theorem and the result of part 8(a) on the previous
H
~ and give the value.
~
page determines C 0 F · ds,
The closed polygonal curve C 0 consists of 5 segments, and is oriented clockwise. The unit
circle C is inside curve C 0 . Let D be the domain between C and C 0 . Then Qx − Py = 0
throughtout D, and Green’s theorem applies. The orientations that keep D on the left
are counterclockwise
on the outside curve C 0 and clockwise on the inner curve C. So
RR
integrals when their
0=
D (Qx − Py ) dA gives the sum of the two boundary circulation H
0
~ = 2π when
orientations are opposite. Since C is oriented clockwise here and since C F~ · ds
H
H
~ So 0 F~ · ds
~ = −2π.
C is oriented counterclockwise, we get 0 = 2π + C 0 F~ · ds.
C
8(c)[3]. If C 00 is the square in the first Hquadrant x > 0, y > 0 with vertices (1, 1), (2, 1), (2, 2), (1, 2)
~ and explain your answer. [Hint: Use any
oriented counterclockwise, evaluate C 00 F~ · ds
method. Reread the first paragraph on the previous page.]
The square encloses a region R in the first quadrant, where Qx − Py = 0 at all points of
R
RR
~ =
R. Therefore, C 00 F~ · ds
R (Qx − Py ) dA = 0 by Green’s Theorem.
10
9(a)[3]. Consider the plane P with equation 2x + y − 2z = 4. Find a unit normal vector ~n
to plane P that points ‘up’, that is, with positive z-coordinate.
A normal vector to a plane ax + by + cz = d is (a, b, c). In our case this gives (2, 1, −2). To
point ‘up’, we ned the z-coordinate to be positive, and we need a unit vector, so we must
divide by the length. Answer: 31 (−2, −1, 2).
(b)[2]. Consider a simple closed curve C on plane P that bounds a region D on plane P .
Choose the unit normal ~n to D as in part (a). If C is oriented with unit tangent vectors
~t so as to be consistent with the conditions of Stokes theorem for C and D, should vector
~t × ~n point into or out of the region D on plane P ? Explain. [Be careful.]
The system for orientations for the purposes of Stokes theorem is that the normal, followed
by the tangent, followed by a vector in the plane pointing into D, should be a right hand
system. This is the same as saying ~n × ~t should point into D. Therefore ~t × ~n should point
out of D on plane P .
(c)[5]. Let F~ = (az, bx, cy), where a, b, c are constants. Calculate curl(F~ ). With orientations ~n and ~t chosen as in parts (a) and (b), so that Stokes theorem
applies, find necessary
H
and sufficient conditions on constants a, b, c that guarantee C F~ · ~t ds > 0.


~k
~i
~j
curl(F~ ) =  ∂/dx ∂/dy ∂/dz  = (c, a, b)
az
bx
cy
Therefore, by Stokes,
I
Z Z
Z Z
1
area(D)
~
~
~
F ·t ds =
curl(F )·~n dS =
(c, a, b)·(−2, −1, 2) dA =
(−2c−a+2b)
3
3
C
D
D
This last expression is positive if and only if −2c − a + 2b > 0. Equivalently, 2b >
2c + a.
11
10(a)[6]. Which of the regions described below are simply connected? Answer YES or NO.
Score is Right minus 12 Wrong, with a minimum of 0. There are six regions.
(i). R3 with the y axis removed. NO. Draw a circle around the y-axis. Any surface
bounding this circle must intersect the y-axis.
(ii). R3 with the circle x2 + y 2 = 1, z = 0 in the xy-plane removed. NO. For example,
draw a circle in the yz-plane of radius 1 centered at (0, 1, 0). Any surface bounding this
circle must intersect the original unit circle in the xy-plane.
(iii). R3 with the origin removed. YES. Any simple closed curve in this region bounds
some surface that misses the origin, hence the surface is also in the region.
(iv). R2 with the origin removed. NO. Consider the unit circle. It’s inside contains the
origin.
(v). The half plane x ≥ 0 in R2 . YES. For any simple closed curve in this region, it’s
inside is still in this region.
(vi). The inside of a square in R2 described as all (x, y) with 0 < x < 1, 0 < y < 1. YES.
Similar to (v).
12
(b)[6]. Write T (true) or F (false) in the margin to the left of each statement below. Score
is Right minus 12 Wrong, with a minimum of 0. There are six statements.
(i). If curl(F~ ) = ~0 on all of R3 , then
H
C
~ = 0 for all closed curves C in R3 .
F~ · ds
TRUE. Since R3 is simply connected, curl(F~ ) = ~0 implies F~ satisfies path independence.
3
The
proof isR via
R Stokes theorem: Any closed curve C bounds a surface S in R . Then
H
~
~ ~
~
S curl(F ) · dS = 0
C F · ds =
~ on R3 with curl(G)
~ = (y, x, z).
(ii). There exists a vector field G
~
FALSE. div(y, x, z) = 1, so this vector field cannot be of form curl(G).
(iii). If F~ is a differentiable vector field on all of R3 and if div(F~ ) = 1 at all
R Rpoints of a
~ n dS
solid region W in R3 and if S is the boundary of W then the outward flux
S F ·~
always equals the surface area of S.
RRR
RR
~ ~
FALSE.
S F · dS =
W 1 dV = V olume(W ).
(iv). If F~ is a differentiable vector field on all of R3 and if
closed curves C in R3 , then F~ = ~0.
H
C
~ = 0 for all simple
F~ · ds
~ would have this property.
FALSE. Any gradient field F~ = ∇f
(v). If F~ is a twice differentiable vector field on R3 −(0, 0, 0), that
R Ris, everywhere except the
~ ~
origin, and if S is the sphere of radius 1 and center (0, 0, 0), then
S curl(F )·dS = 0.
TRUE. The flux integral of any curl across any closed surface is 0, as a consequence of
Stokes theorem.
3
(vi). If F~ is a conservative vector field, differentiable
R R on all of R , and if S is a closed
~
~
oriented surface, that is, with no boundary, then
S F · dS = 0.
RR
RRR
~ =
~
~
FALSE.
S ∇f · dS
R div(∇f ) dV , where S bounds solid region R. But
~ ) = ∂ 2 f /∂x2 + ∂ 2 f /∂y 2 + ∂ 2 f /∂z 2 can be pretty much anything. E.g. if you start
div(∇f
with f = x2 , then the Laplacian is 2.
13
11(a)[4]. Let f (x, y, z) be a function on R3 and F~ (x, y, z) a vector field on R3 , both with
continuous partial derivatives everywhere. There is a ‘product rule’ for div(f F~ ), expressing
this divergence as a sum of two terms. Prove this product rule.
[You don’t need to be told the rule. Just work out div(f F~ ) yourself and see the rule. As
a hint, one of the two terms is a dot product.]
Write F~ = (P, Q, R). Then div(f F~ ) =
∂
∂
∂
∂x (f P ) + ∂y (f Q) + ∂z (f R)
= (fx P + f Px ) + (fy Q +
f Qy ) + (fz R + f Rz ). The first, third and fifth terms add to grad(f ) · F~ . The second, fourth
and sixth terms add to f div(F~ ). So we have proved
~ · F~ + f div(F~ )
div(f F~ ) = ∇f
[A lot of students had nonsense like ∂ F~ /∂x all over the place. You have to work with the
three component functions of F~ separately.]
(b)[4]. If F~ = (P, Q, R) is a twice differentiable vector field in a region W in R3 , show that
div(curl(F~ )) = 0 in W . Be clear about how the hypotheses justify your work.
curl(F~ ) = (Ry − Qz , Pz − Rx , Qx − Py ). Computing the divergence gives a sum of six
terms, all of them second partial derivatives of P, Q, or R. Since P, Q, R are assumed to
have continuous second partial derivative, ”mixed partials” are equal, e.g. ∂ 2 P/∂y∂z =
∂ 2 P/∂z∂y, etc. Three such pairs occur in the above six terms with opposite signs. It is
thus seen that everything cancels out in the divergence calculation, giving 0.
14
12[10]. The graph on the next two pages ‘plots’ a vector field F~ (x, y) = (P (x, y), Q(x, y)),
oriented curves A, B, C, and points M, N, R, T . Answer the following ten questions. There
are two copies of the graph so you can tear off the last page while contemplating the
questions. Score is, as usual, Right − 12 Wrong, with a minimum of 0. [One assumes F~ is
differentiable.]
The point here is to carefully look at dot products F~ · ~t and F~ · ~n at various points in
the picture, like around curves, or near specified points. ~t’s must always point in the
specified orientation directions of drawn curves, or counterclockwise when encircling one
point. Acute angles between vectors give positive dot products, obtuse angles give negative
dot products. All the integrals amount to summing such dot products multiplied by small
arc lengths, around curves.
(i). Is the circulation of F~ around curve B positive, negative, or 0?
NEGATIVE. F~ and tangents ~t in the direction given make obtuse angles.
(ii). If a unit normal ~n is chosenRto curve C pointing down through the letter C at points
near that letter, is the quantity C F~ · ~n ds positive, negative, or 0?
POSITIVE. F~ and the ‘downward’ normals ~n to C make some acute angles and maybe
some right angles.
(iii). Is the quantity
H
A
~ positive, negative, or 0?
F~ · ds
POSITIVE. Tangents to directed curve A make acute angles with F~ .
(iv). Is the value of
dQ
dx
−
dP
dy
at T positive, negative, or 0?
NEGATIVE. F~ makes obtuse angles with counterclockwise ~t on small circles around
T.
(v). Is the quantity
dP
dx
+
dQ
dy
at R positive, negative, or 0?
NEGATIVE. Flow F~ is INTO small regions around R. That is, F~ and outward normals
to small circles around R make obtuse angles.
(vi). Is the divergence of F~ at M positive, negative, or 0?
POSITIVE. Flow is away from point M .
15
(vii). If D is the plane region bounded by curve A, is the quantity
positive, negative, or 0?
RR
D
div(F~ ) dxdy
NEGATIVE. Flow is into the region bounded by A. Outward normals to A are obtuse to
F~ .
(viii). If F~ is a fluid flow velocity field, would a small paddle at N spin clockwise, counterclockwise, or not at all?
COUNTERCLOCKWISE. The circulation at N is positive, flow vectors F~ are acute with
counterclockwise tangents to small circles at N .
(ix). If F~ is a force field, is the total work done by F~ on a particle moving along curve C
positive, negative, or 0?
POSITIVE. Some angles look perpendicular, but on the second half of C, F~ vectors are
acute with the directed tangents of C.
~ for some function f (x, y)?
(x). Is it possible that F~ = ∇f
OF COURSE NOT. Many of the conclusions above amounted to curl(F~ ) 6= 0 at a point,
or to a non-zero vector line integral of F~ around closed curves.
16
y
T
R
B
A
M
N
C
x
17