MATH 52 FINAL EXAM SOLUTIONS
1. (a) Sketch the region R of integration in the following double integral.
Z 1Z 1
5
xey dy dx
√
x
0
√
Solution. R = {(x, y) | 0 ≤ x ≤ 1, x ≤ y ≤ 1}.
(b) Express the region R as an x-simple region.
Solution. R = {(x, y) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y 2}
(c) Evaluate the integral by changing the order of integration.
Solution.
x=y2
Z 1
Z 1 Z y2
1 2 y5
y5
dy
xe
xe dx dy =
2
0
0
0
x=0
1
Z 1
1 4 y5
1 y5
=
y e dy =
e
10
0 2
0
e−1
=
10
2. (a) Let T be a solid cone whose base is the disk x2 + y 2 ≤ 1 in the xy-plane and
whose vertex is the point (0, 0, 2). Set up, but do not evaluate, a triple integral
in cylindrical coordinates which computes the moment of inertia of T around the
x-axis.
Solution. In cylindrical coordinates,
T = {(r, θ, z) | 0 ≤ z ≤ 2 − 2r, 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π},
so
Ix =
ZZZ
2
2
y + z dV =
T
Z
0
2π
Z
0
1
Z
2−2r
(r 2 sin2 θ + z 2 )r dz dr dθ
0
(b) Let T be the solid ball of radius 1 centered at (2, 0, 0). Set up, but do not evaluate,
a triple integral in spherical coordinates which computes the moment of inertia
of T around the z-axis. Hint: First consider the change of variables u = x − 2,
v = y, w = z.
Solution. After the change of variables, the sphere takes the form u2 +v 2 +w 2 = 1,
and the z-axis becomes the line (u, v) = (−2, 0). This means the distance from
the z-axis is (u + 2)2 + v 2 , and therefore
Iz =
Z
0
2π
Z
0
π
Z
1
[(ρ cos θ sin φ + 2)2 + (ρ sin θ sin φ)2 ]ρ2 sin φ dρ dφ dθ
0
3. Consider the curve C parametrized by r(t) = (2et , 31 e3t + e−t ) for 0 ≤ t ≤ 1.
1
(a) Find the arclength of C.
Solution. Since r0 (t) = (2et , e3t − e−t ),
Z 1√
Z
Z 1
0
kr (t)k dt =
s=
1 ds =
4e2t + e6t − 2e2t + e−2t dt
0
0
C
1
Z 1
1 3t
3t
−t
−t
e + e dt = e − e
=
3
0
0
1 3
2
= e − e−1 +
3
3
(b) Find the x-coordinate of the centroid of C.
Solution.
Z
Z
1
1 1 t 3t
x̄ =
x ds =
2e (e + e−t ) dt
s C
s 0
1
Z
1 1 4t
1 1 4t
2e + 2 dt =
=
e + 2t
s 0
s 2
0
1 1 4 3
e +
=
s 2
2
(c) Let S be the surface obtained by rotating C around the line x = −1. Find the
area of S.
Solution.
By Pappus’ Theorem, area(S) = s · d, where d is the distance travelled by the centroid.
Now the centroid travels along a circle of radius 1 + x̄, hence d = 2π(1 + x̄). So
area(S) = 2πs(1 + x̄) = 2π( 31 e3 − e−1 + 32 + 12 e4 + 23 )
4. (a) Consider the change of variables u =
x2
4
+ y 2 , v = xy . Find the Jacobian
∂(x, y)
∂(u, v)
Solution.
Hence
∂(u, v) 12 x − xy2 1
y2
1
1
=
+
2
=
= + 2v 2 = (1 + 4v 2)
1 2
2y
∂(x, y)
2
x
2
2
x
∂(x, y)
2
=
∂(u, v)
1 + 4v 2
2
2
(b) Find the area of the region in the first quadrant enclosed by the ellipses x4 +y 2 = 1,
x2
+ y 2 = 4 and the lines y = 2x and y = 4x.
4
Solution.
The area is therefore
Z 4
Z 4Z 4
2
1
dv du = 6
dv
A=
2
2
2 1 + 4v
v=2 1 + 4v
1
Z 8
1
=3
dw = [3 arctan(w)]84 = 3(arctan(8) − arctan(4))
2
1
+
w
w=4
5. Let T be the solid enclosed by the planes z = x + y, x + 2y = 2 and the paraboloid
z = y 2 . Set up, but do not evaluate a triple integral in rectangular coordinates which
computes the volume of T by
(a) regarding T as x-simple.
Solution.
If we regard T as x-simple we have a cylinder z = y 2 and two graphs x = z − y
and x = 2 − 2y. Now the two graphs intersect along a line z + y = 2. This line
intersects the parabola z = y 2 in two points: (y, z) = (1, 1) and (y, z) = (−2, 4).
So for T the coordinate ranges are as follows: −2 ≤ y ≤ 1, y 2 ≤ z ≤ 2 − y and
z − y ≤ x ≤ 2 − 2y. So
Z 1 Z 2−y Z 2−2y
vol(T ) =
1 dx dz dy
y=−2
z=y 2
x=z−y
(b) regarding T as z-simple.
Solution.
If we regard T as z-simple we have a cylinder x + 2y = 2 and two graphs z = x + y
and z = y 2. Now the graphs intersect along a parabola x = y 2 − y. This parabola
intersects a line x + 2y = 2 in two points: (x, y) = (0, 1) and (x, y) = (6, −2). So
for T the coordinate ranges are as follows: −2 ≤ y ≤ 1, y 2 − y ≤ x ≤ 2 − 2y and
y 2 ≤ z ≤ x + y. So
Z 1 Z 2−2y Z x+y
vol(T ) =
1 dx dz dy
y=−2
x=y 2 −y
z=y 2
6. Let C be the curve parametrized by r(t) = (t2 , sin t) for 0 ≤ t ≤ π/2.
Z
(a) Compute
y dx.
C
Solution. Since r0 (t) = (2t, cos t), we have
Z
C
y dx =
Z
π/2
2t sin t dt =
0
3
π/2
=2
− 2t cos t + 2 sin t
0
(b) Let R be the region in the plane bounded by the y-axis, the line y = 1, and the
curve C. Use Green’s Theorem to find the area of R.
Solution. By Green’s Theorem,
I
ZZ
y dx =
−1 dA = − area(R)
∂R
R
Now the boundary of R consists of the curve C, followed by the line segment
L1 from (π 2 /4, 1) to (0, 1) and the line segment L2 from (0, 0) to (0, 0). These
segments are parametrized by (π 2 /4(1 − t), 1) and (0, 1 − t), respectively, for
0 ≤ t ≤ 1. Thus
Z
I
Z
Z
y dx
y dx +
y dx =
y dx +
L2
∂R
C
L1
Z 1
Z 1
2
= 2+
−π /4 dt +
0 dt = 2 − π 2 /4
0
0
which implies that area(R) = π 2 /4 − 2.
7. Let S be the portion of the cylinder x2 + y 2 = 2x which lies above the xy-plane and
below the surface z = x2 .
(a) Write down a parametrization of S. Be sure to specify the domain.
Solution. There are several possible solutions. Here are four of them.
• First parametrize the circle x2 + y 2 = 2x by x = 1 + cos t, y = sin t for
0 ≤ t ≤ 2π. Then the z-coordinate varies from 0 to x2 = (1 + cos t)2 . So
r(t, z) = (1 + cos t, sin t, z)
0 ≤ t ≤ 2π, 0 ≤ z ≤ (1 + cos t)2 .
• A variant of the first solution is to let z = u(1 + cos t)2 with u varying from
0 to 1:
r(t, u) = (1 + cos t, sin t, u(1 + cos t)2 )
0 ≤ t ≤ 2π, 0 ≤ u ≤ 1.
• Using polar coordinates to describe the circle, we have r = 2 cos θ for −π/2 ≤
θ ≤ π/2, so x = r cos θ = 2 cos2 θ and y = r sin θ = 2 cos θ sin θ. The zcoordinate again varies from 0 to x2 = 4 cos4 θ. So
r(θ, z) = (2 cos2 θ, 2 cos θ sin θ, z)
− π/2 ≤ θ ≤ π/2, 0 ≤ z ≤ 4 cos4 θ.
• A variant of the previous parametrization is
r(θ, u) = (2 cos2 θ, 2 cos θ sin θ, 4u cos4 θ)
− π/2 ≤ θ ≤ π/2, 0 ≤ u ≤ 1.
(b) Find the area of S.
Solution. Using the first parametrization above, we have
rt = (− sin t, cos t, 0)
4
and
rz = (0, 0, 1)
so
krt × rz k = k(cos t, sin t, 0)k = 1
and therefore
area(S) =
Z
0
2π
Z
(1+cos t)2
1 dz dt =
0
Z
2π
(1 + cos t)2 dt = 3π
0
8. Let S be the surface parametrized by r(u, v) = (v 2 , u2,
[−1, 1] × [−1, 1] in the uv-plane.
√
2uv) over the domain D =
(a) Find the area of S.
√
√
Solution. Since ru = (0, 2u, 2v) and rv = (2v, 0, 2u), we have
√
√
√
√
kru × rv k = k(2 2u2 , 2 2v 2 , −4uv)k = 8u4 + 8v 4 + 16u2 v 2 = 2 2(u2 + v 2 ).
Thus
area(S) =
=
ZZ
Z
1 dS =
1
Z
1
kru × rv k du dv
√
√ 2
16 2
2
2 2(u + v ) du dv =
3
−1
S
1 Z 1
−1
Z
−1
−1
(b) Find the x-coordinate of the centroid of S.
Solution. Since x = v 2 on the surface S,
ZZ
Z 1Z 1
√
3
7
1
v 2 · 2 2(u2 + v 2 ) du dv = .
x̄ =
x dS = √
area(S) S
15
16 2 −1 −1
9. Let S be the portion of the sphere x2 + y 2 + z 2 = 4 which lies above the plane z = −1,
and let F(x, y, z) = (x − 3y + 2z, −3x + 2y + z, 2x + y − 3z). Compute the flux
ZZ
F · n dS,
S
where n denotes the outward unit normal to S.
Solution.
√
Let D be the disk in the plane z = −1 centered at (0, 0, −1) with radius 3. The
S ∪ D forms the boundary of the solid region T consisting of the portion of the ball
x2 + y 2 + z 2 ≤ 4 with z ≥ −1. Thus by the Divergence Theorem
ZZ
ZZZ
ZZZ
F · n dS =
div F dV =
0 dV = 0.
S∪D
Now
ZZ
S∪D
T
F · n dS =
ZZ
S
5
T
F · n dS +
ZZ
D
F · n dS,
where in the last integral n = (0, 0, −1). Hence
ZZ
ZZ
ZZ
ZZ
F · n dS = −
F · n dS = −
−2x + y + 3z dS =
2x − y + 3 dS,
S
D
D
D
since z = −1 on D. Now by the symmetry of D and the fact that x and y are odd
functions, this reduces to
ZZ
3 dS = 3 · area(D) = 9π.
D
10. Let C be the triangle with vertices A = (1, 1, 1), B = (3, −1, 0) and C = (1, −1, 2),
parametrized in that order. Let F(x, y, z) = (4z 2 , 2x2 , y 2).
(a) Find the area of the triangle.
Solution. The vectors from A to B and A to C are given by v = (2, −2, −1)
and w = (0, −2, 1), respectively. The area of the triangle is half the area of the
parallelogram spanned by these vectors, so
1
1
area(4ABC) = kv × wk = k(−4, −2, −4)k = 3
2
2
(b) Find the equation of the plane in which the triangle lies.
Solution. From part (a) we know that (2, 1, 2) is a normal vector to the plane, so
using the point (1, 1, 1) we see that the equation of the plane is 2x + y + 2z = 5.
(c) Compute curl F.
Solution. curl F = (2y, 8z, 4x).
I
(d) Compute
F · dr.
C
Solution. Let S denote the portion of the plane bounded by C. Then by Stokes’
Theorem,
I
ZZ
F · dr =
curl F · n dS
C
S
where the direction of n must be the same as v × w from the solution of part (a).
That is, n = (− 32 , − 13 , − 23 ). Using this together with the results of parts (a), (b)
and (c), we get
ZZ
ZZ
ZZ
1
1
20
curl F·n dS = −
4y +8z +8x dS = −
20 dS = − ·area S = −20
3 S
3 S
3
S
11. Let F(x, y, z) = (2xy + z 2 , 2yz + x2 , 2zx + y 2).
(a) Find a potential for F.
Solution. f (x, y, z) = x2 y + z 2 x + y 2 z.
6
(b) Z
Let C be the curve parametrized by r(t) = (cos t, sin t, t) for 0 ≤ t ≤ π. Compute
F · dr
C
Solution. The curve begins at (1, 0, 0) and ends at (−1, 0, π), so by the Fundamental Theorem of Line Integrals,
Z
F · dr = f (−1, 0, π) − f (1, 0, 0) = −π 2 .
C
12. Let T be the solid region which lies below the surfaces z = 1 − x2 and z = 1 − y 2 and
above the xy-plane.
(a) Find the volume of T .
Solution. As a y-simple region, T takes the form
√
√
{(x, y, z) | −1 ≤ x ≤ 1, 0 ≤ z ≤ 1 − x2 , − 1 − z ≤ y ≤ 1 − z}.
Using the symmetry of the region, we have
vol(T ) = 4
Z 1Z
0
1−x2
0
Z
√
1−z
1 dy dz dx = 4
0
Z 1Z
0
0
Z 1
2
− (1 − z)3/2
=4
3
0
1
8
1 4
=
x− x
=2
3
4
0
1−x2
dx = 4
Z
0
0
1
1−x2
√
1 − z dz dx
2
1 − x3 dx
3
(b) Find the outward flux through the boundary of T of the vector field F(x, y, z) =
(zx2 , z 2 − y, 3z − z 2 x).
Solution. By the Divergence Theorem,
ZZ
ZZZ
ZZZ
F · n dS =
div F dV =
2 dV = 2 vol(T ) = 4
∂T
T
T
7