Calculus III
Math 233 — Spring 2007
In-term exam April 11th. — Suggested solutions
This exam contains sixteen problems numbered 1 through 16. Problems 1 – 15 are multiple
choice problems, which each count 5% of your total score. Problem 16 will be hand-graded
and counts 25% of your total score.
Problem 1
Evaluate
A) π
R 11 R 33
4
2
y sin(πx) dx dy.
B) 31π
C)
14
π
D)
105
π
E) 0
F)
1
π
G) 14
H) 7π
This is a straightforward calculation,
x=33
Z 11 Z 11 Z 33
Z 11 1
1
dy =
− y (−1) − 1 dy
y sin(πx) dx dy =
− y cos(πx)
π
π
4
4
2
4
x=2
Z 11
1 11 105
1
=
2y dy = y 2 4 =
.
π 4
π
π
1
Problem 2
Use Lagrange multipliers to find the maximum value of the function
f (x, y, z) = x − y + z
on the sphere (x − 1)2 + (y − 2)2 + (z − 1)2 = 4.
√
√
√
√
4
A)
3
B) 23 3
C) 41 3
D) 1 + 23 3
3 √ √
G) 2 3
H) 2 − 23 3
E) 1
F)
1
6
√
3
Let g(x, y, z) = (x − 1)2 + (y − 2)2 + (z − 1)2 − 4. Then the method of Lagrange multipliers
~ = λ∇g.
~ Because ∇f
~ = h1, −1, 1i
says that a maximum can only occur at a point where ∇f
~
and ∇g = h2(x − 1), 2(y − 2), 2(z − 1)i, we need to solve

1 = 2λ(x − 1)



−1 = 2λ(y − 2)
1 = 2λ(z − 1)



4 = (x − 1)2 + (y − 2)2 + (z − 1)2 .
The first three equations tell us that x − 1 = −(y − 2) = z − 1. By the last equation,
3(x − 1)2 = 4, or x − 1 = ± √23 . Because y = 3 − x and z = x, the value of f is
f (x, y, z) = x − y + z = x − (3 − x) + x = 3 x − 1 .
The maximum value is therefore attained when x − 1 =
2
√2 ,
3
where it is 3 ·
√2
3
√
= 2 3.
Problem 3
Find the surface area of the part of z = 1 − x2 − y 2 that lies above the xy-plane.
A)
5π
6
√
√
G) 2π 7
B)
3π
4
H)
C)
√
2
π
D) 1 +
2√
π
3
E)
√π
5
2 3−1
3
F)
√
(5 5−1)π
6
This is about the surface area of a graph, so we can use formula 6 on page 870 of our
calculus book. We need the partial derivatives of z:
∂z
= −2x,
∂x
∂z
= −2y.
∂y
The function we need to integrate, is
p
p
1 + (−2x)2 + (−2y)2 = 1 + 4(x2 + y 2 ).
Then we need to find the domain: since z = 1 − x2 − y 2 ≥ 0, we get x2 + y 2 ≤ 1. So it
looks like it would be best to use polar coordinates. Now we can set up the integral,
Z 2π Z 1 √
1 + 4r2 r dr dθ
0
0
This calls for a substitution. Defining u = 1 + 4r2 , we get du = 8r dr, so r dr =
Hence,
Z 2π Z 1 √
Z 2π Z 5
Z 2π
√
1
1 2 3/2 u=5
2
u
dθ
1 + 4r r dr dθ =
u du dθ =
8
8 3
u=1
0
0
0
1
0
Z 2π
√
√
1
1
5
5
5
−
1
dθ
=
5
−
1
π.
=
12
6
0
3
1
8
du.
Problem 4
Find the volume of the solid bounded by z = 1 − x2 − y 2 and the xy-plane.
A)
1
2
B)
2
3
C) 1
D)
4
3
E)
π
2
F)
2π
3
G) π
H)
4π
3
To find the volume we will integrate the height of the solid, over the projection of the solid
in the xy-plane. The projection of the solid is bounded by the circle x2 + y 2 = 1, while the
height is 1 − x2 − y 2 . Everything is conveniently described using polar coordinates, so we
calculate
Z 2π Z 1
Z 2π Z 1
2
(1 − r ) · r dr dθ =
r − r3 dr dθ
V =
0
Z0 2π 0
Z 2π 0
1 2 1 4 1
dθ = π2 .
=
r − 4 r 0 dθ = 41
2
0
0
4
Problem 5
Let D be the triangle with vertices (0, 2), (3, −1), and (3, 2). Calculate
ZZ
xy dA.
D
A)
63
8
B)
37
4
C)
57
2
51
8
D)
E)
41
2
F)
33
4
G)
47
6
H)
11
4
First, let’s draw a picture of our triangle D:
3
2
1
0
−1
0
1
2
3
4
−1
−2
This can be seen as either a Type I or a Type II region. As a Type I region, it produces
the following integral,
ZZ
Z 3Z 2
Z 3
Z 3
1 2 y=2
1
xy dA =
xy dy dx =
xy y=2−x dx =
x 4 − (2 − x)2 dx
2
2
D
0
2−x
0
0
Z 3
Z 3
3
1
x 4x − x2 dx =
= 63
.
=
2x2 − 21 x3 dx = 23 x3 − 18 x4 0 = 18 − 81
2
8
8
0
0
Problem 6
Find the Jacobian
∂(x,y)
∂(u,v)
of the transformation
x = 3u + v,
A) −10
B) −5
C) −2
y = 4u − 2v.
D) 0
E) 1
F) 3
The Jacobian is given by
∂(x, y)
∂x ∂y ∂x ∂y
=
−
= 3 · (−2) − 1 · 4 = −10.
∂(u, v)
∂u ∂v ∂v ∂u
5
G) 4
H) 6
Problem 7
Compute the volume of the pyramid with a rectangular base that is bounded by the
five planes 3x + z = 12, −3x + z = 12, 3y + z = 12, −3y + z = 12 and z = 0.
Hint: Use symmetry to calculate the volume of the pyramid as four times the volume
of a tetrahedron.
A) 64
H) 432
B) 84
C) 128
D) 172
E) 216
F) 256
G) 342
If we project the pyramid onto the xy-plane, we get the following:
4
2
0
−4
−2
0
2
4
−2
−4
For instance, the plane 3x + z = 12 intersects the xy-plane in the line x = 4 (we find this
by substituting z = 0 in the equation for the plane). We can calculate the total volume
as four times the volume of one of the tetrahedrons of which the pyramids consists, for
example the one with vertices (4, 4, 0), (4, −4, 0), (0, 0, 0) and (0, 0, 12). This one is in fact
the solid between D and the plane 3x + z = 12, where D is the triangle in the xy-plane
with vertices (4, 4), (4, −4) and (0, 0). This region D is a Type I region, so we can set up
the integral to compute the volume:
ZZ
Z 4Z x
Z 4
y=x
(12 − 3x) dy dx =
(12 − 3x)y y=−x dx
(12 − 3x) dA =
0
−x
0
D
Z 4
Z 4
4
24x − 6x2 dx = 12x2 − 2x3 0
=
(12 − 3x) · 2x dx =
0
0
= 192 − 128 = 64.
So the volume of the pyramid is 4 · 64 = 256.
6
Problem 8
Evaluate
RRR
1
E x2 +y 2 +z 2
dV , where E is the solid bounded by two spheres:
E = {(x, y, z) : 1 ≤ x2 + y 2 + z 2 ≤ 4}.
A) 28π
3
H) π
B) 4π
C) 23 π
D)
4π
3
E)
32π
3
F) 3π
G) 2π
In spherical coordinates E is described by 1 ≤ ρ ≤ 2, 0 ≤ φ ≤ π, and 0 ≤ θ ≤ 2π. The
integrand x2 +y12 +z2 is written ρ12 in spherical coordinates. We integrate
ZZZ
E
Z 2π Z π Z 2
1 2
ρ sin φ dρ dφ dθ =
sin φ dρ dφ dθ
2
0
0
0
0
1
1 ρ
Z 2π Z π
Z 2π Z π
=
(2 − 1) sin φ dφ dθ =
sin φ dφ dθ
0
0
0
0
Z 2π
Z 2π
φ=π
2 dθ = 2(2π − 0) = 4π.
=
− cos φ φ=0 dθ =
1
dV =
2
x + y2 + z2
Z
2π
Z
π
Z
2
0
0
7
Problem 9
Which of the following polar equations defines the circle below?
1.0
0.75
0.5
0.25
0.0
−1.0
−0.75
−0.5
−0.25
0.0
A) r = 1 − 2 sin(θ)
B) r = cos(θ)
C) r = sin(θ) − cos(θ)
D) r = θ cos(θ)
E) r = θ sin(2θ)
F) r = 2 + sin(2θ)
G) r = sin(θ) − 1
H) r = θ
There are several ways to solve this problem. We will provide you with two of them:
1. The point (x, y) = (0, 1) is on the circle. In polar coordinates this point can be
written (r, θ) = (1, π2 ) or (r, θ) = (−1, 3π
). The equation in answer C) is the only
2
one which this point satisfies.
q
√
1 1
1
2. The circle in the picture has center (− 2 , 2 ) and radius 2 2 = 12 . So the equation
in terms of x and y is
(x + 21 )2 + (y − 12 )2 = 21 .
Rewriting to polar coordinates yields
(x + 12 )2 + (y − 12 )2 =
1
2
Convert to polar coordinates
8
⇔
⇔
⇔
x2 + x + 41 + y 2 − y + 14 =
x2 + y 2 = y − x
r2 = r sin(θ) − r cos(θ)
⇔
r = sin(θ) − cos(θ).
1
2
Problem 10
Evaluate
1
Z
0
Z
1
√
cos y 3 dy dx.
x
by changing the order of integration.
B) 31 cos(1)
C) 13 sin(1) − 1
A) 13 cos(1) − 1
D)
E) cos(1) − 1
F) cos(1)
G) sin(1) − 1
H) sin(1)
1
3
sin(1)
To change the order of integration, we first draw the region we are integrating over.
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.25
0.5
1.0
0.75
x
Here we see that if we want to integrate with respect to x first, we need to let x run between
0 and y 2 , for y between 0 and 1. We can then calculate the integral, using the substitution
u = y3,
Z
0
1
Z
1
√
cos y
3
Z
1
Z
dy dx =
x
y2
cos y
0
Z
=
0
3
0
1
Z
1
dx dy =
(y 2 − 0) cos y 3 dy
0
1
1
1
cos
u
du
=
sin
u
=
3
3
0
9
1
3
sin(1) − 0 = 13 sin(1).
Problem 11
Find the tangent vector of the curve with parametric equations
Z t
Z t
2
1
x(t) =
sin 2 πθ dθ,
and
y(t) =
cos 12 πθ2 dθ
0
0
at the point corresponding to t = 1.
√ √
√ √
A) h1, 0i
B) h 21 2, 12 2i
C) h0, 1i
D) h− 21 2, 21 2i
√ √
√
√
F) h− 12 2, − 21 2i
G) h0, −1i
H) h 21 2, − 21 2i
E) h−1, 0i
The Fundamental Theorem of Calculus tells us that
r~0 (t) = hx0 (t), y 0 (t)i = hsin( 21 πt2 ), cos( 12 πt2 )i
and substituting t = 1 gives
r~0 (1) = hsin( 12 π), cos( 21 π)i = h1, 0i.
Problem 12
Describe the boundary of the solid region E that we integrate over in
ZZZ
Z
1
f (x, y, z) dV =
−1
E
√
Z
1−x2
√
− 1−x2
Z
2−x2 −y 2
√
f (x, y, z) dz dy dx.
x2 +y 2
A) A paraboloid and a plane
B) A sphere
C) A cone and a half-sphere
D) A paraboloid and
a
half-sphere
E)
A
cone
and
a plane
F) An ellipsoid
H) A cone and a paraboloid
G) Two cones
The region E is described by
p
√
√
E = {(x, y, z) : x2 + y 2 ≤ z ≤ 2 − x2 − y 2 , − 1 − x2 ≤ y ≤ 1 − x2 , −1 ≤ x ≤ 1}.
p
The boundary of E is described by the surfaces z =
x2 + y 2 , which is a cone, and
2
2
z = 2 − x − y , which is a paraboloid.
10
Problem 13
Let D be the region bounded by the circle x2 + y 2 = 2x. Evaluate
ZZ
x2 + y 2 y dA.
D
A) 0
B)
1
30
C)
1
20
D)
1
10
E)
π
10
F)
π
2
G) π
H) 2π
The expression ”x2 + y 2 ” appears twice in this problem, so it would probably be best to
use polar coordinates. First of all, let’s rewrite the equation of the circle:
x2 + y 2 = 2x
x2 − 2x + 1 + y 2 = 1
⇔
(x − 1)2 + y 2 = 1.
⇔
So the circle has center (1, 0) and radius 1.
1.5
1.0
0.5
0.0
−0.5
0.0
−0.5
1.0
0.5
2.0
1.5
2.5
−1.0
−1.5
Let us rewrite the equation in terms of polar coordinates:
x2 + y 2 = 2x
⇔
r2 = 2r cos θ
⇔
r = 2 cos θ.
There are different possible values for the bounds of r and θ for the region within this
circle, but the most convenient is − π2 ≤ θ ≤ π2 and 0 ≤ r ≤ 2 cos θ. We can evaluate the
integral:
Z π Z 2 cos θ
ZZ
Z π Z 2 cos θ
2
2
2
2
2
(x + y )y dA =
r · r sin θ · r dr dθ =
r4 sin θ dr dθ
−π2
D
Z
π
2
=
−π2
−π2
0
r5 sin θ
5
1
r=2 cos θ
r=0
11
Z
π
2
dθ =
−π2
32
5
0
cos5 θ sin θ dθ.
If we now do the substitution u = cos θ, then du = − sin θ dθ, whence
π
2
Z
32
5
−π2
Z
5
0
32 6 0
5
u
du
=
− 30 u 0 = 0.
− 32
5
cos θ sin θ dθ =
0
Here’s another way to see it:
Z
π
2
−π2
32
5
π2
6
cos5 θ sin θ dθ = − 32
cos
θ
= 0 − 0 = 0.
30
−π
2
Alternatively, we can just argue that
must be equal to zero.
32
5
cos5 θ sin θ is an odd function of θ, so the integral
Problem 14
√
Let S be the surface of revolution obtained by rotating the curve y = 2 x, 3 ≤ x ≤ 8,
about the x-axis. Compute the surface area of S.
A)
G)
152π
3
251π
3
B) 523π
H)
C)
231π
5
D)
315π
2
E)
132π
5
F) 325π
513π
2
To calculate the surface area of a surface of revolution, we can use the formula
Z b
p
area(S) = 2π
f (x) 1 + [f 0 (x)]2 dx,
a
or the more general
ZZ
|~rx × ~rθ | dA.
area(S) =
D
In the latter case, we use the vector function ~r(x, θ) = hx, f (x) cos θ, f (x) sin θi.
Let us use the first formula. We calculate the derivative of f , f 0 (x) =
Z 8
√
√ q
1
area(S) = 2π
2 x 1 + x dx = 4π
x + 1 dx
3
3
√
8
= 4π 23 (x + 1) x + 1 3 = 8π
9·3−4·2 =
3
Z
2
√
2 x
8
12
152π
.
3
=
√1 .
x
Then
Problem 15
Which vector field do you see in the picture ?
3
2
y
1
0
−3
−2
0
−1
1
2
3
x
−1
−2
−3
A) F~ (x, y) = h x, y i
B) F~ (x, y) = h −x, y i
C) F~ (x, y) = h x, −y i
D) F~ (x, y) = h −x, −y i
E) F~ (x, y) = h y, x i
F) F~ (x, y) = h −y, x i
G) F~ (x, y) = h y, −x i
H) F~ (x, y) = h −y, −x i
On the positive x-axis (where y = 0), the arrows are horizontal and pointing in the positive
direction of x. Using this argument, and looking at the formulas for the vector fields in
the eight answers, the only possible correct answers are A) and C).
On the positive y-axis (where x = 0), the arrows are vertical and pointing in the negative
y-direction. This leaves C) as the correct answer.
13
Name:
9 – 10 Hjelle
Section: 10 – 11 Berson
11 – 12 Hjelle
Student-ID:
The following problem will be hand-graded. To earn full credit you need to justify your
answers.
Problem 16
a) Calculate
Z
3
−3
√
Z
−
9−x2
√
3−
p
x2 + y 2 dy dx.
9−x2
b) Find the volume of the solid bounded by the cone z =
x2 + y 2 = 9 and the xy-plane.
p
x2 + y 2 , the cylinder
a) We recognize that the integral is over the disk D with center at the origin and radius
3. Also, the integrand can be expressed in polar coordinates as 3 − r. Therefore
Z
3
−3
√
Z
9−x2
√
− 9−x2
3−
p
2π
Z
x2
+
y2
dy dx. =
0
=
27
2
2π 3 2
1 3 3
(3 − r) r dr dθ =
r
−
r
dθ
2
3
0
0
0
Z
2π
dθ = 92 (2π − 0) = 9π.
−9
Z
3
Z
0
b) To find the volume, we need to integrate the height of the cone over the projection
of the solid to the xy-plane. Since the solid lies inside the cylinder x2 + y 2 = 9, the
projection
is the disk with center at the origin and radius 3. The height is given by
p
z = x2 + y 2 = r. Thus,
Z 2π Z 3
V =
r · r dr dθ.
0
0
This is part of the integral we calculated above, so
Z 2π
V =9
dθ = 18π.
0
Alternatively, we could have seen that in a), we calculated the volume of the given
cylinder cut off at height 0 and height 3, with everything under the cone removed.
Since the volume of the cylinder is V = π · h · r2 = π · 3 · 32 = 27π, the volume under
the cone is 27π − 9π = 18π.