Math 221
Final Exam Sample Problems
Answers
December 4, 2015
1. Since ρ2 = z 2 + r2 , we see that ρ2 = 4 + 12 = 16, and so ρ = 4. Also z = ρ cos φ, so
cos φ = 24 = 21 , which implies that φ = π3 . Finally, θ is unchanged between cylindrical
and spherical coordinates, so the answer is (ρ, θ, φ) = (4, π3 , π3 ) .
2. By setting 5 = 9 − x2 − y 2 , we see that the curve of intersection projects onto the circle
x2 + y 2 = 4 in the xy-plane. In the first quadrant (which sits below
p the first octant)
this projects down to the quarter disk 0 ≤ y ≤ 2 and 0 ≤ x ≤ 4 − y 2 , which then
gives the bounds for x- and y-coordinates of points in E. Finally the z-coordinate of
a point in E is bounded by 5 ≤ z ≤ 9 − x2 − y 2 , so the volume of E is
2
Z
ZZZ
Vol(E) =
Z √4−y2 Z
9−x2 −y 2
dz dx dy .
dV =
0
E
5
0
3. The region E is the snow cone shaped region between the cone and the sphere.
(a) We calculate the bounds on E in spherical coordinates. First 0 ≤ θ ≤ 2π, since
we are rotating completely around the z-axis. The cone z 2 = x2 + y 2 determines the
range of φ. Since z = ρ cos φ and r = ρ sin φ, we see that on this cone, where z 2 = r2 ,
we have tan φ = 1, which means the cone is described by the equation
π
.
4
φ=
Thus the φ range of E is 0 ≤ φ ≤
√
sphere, 18. Thus
π
.
4
ZZZ
Finally, ρ ranges from 0 to the radius of the
Z
Vol(E) =
2π
Z
π
4
√
Z
18
ρ2 sin φ dρ dφ dθ.
dV =
E
0
0
0
(b) To convert the limits of integration on E into cylindrical coordinates, we first see
that E is contained in the cylinder x2 + y 2 = 9 (obtained by substituting z 2 = x2 + y 2
into x2 + y 2 + z 2 = 18). Thus E sits over the disk of radius 3 centered at the origin
and is bounded above by the sphere and below by the cone. In cylindrical coordinates,
the sphere is given by z 2 + r2 = 18 and the cone is given by z = r. The mass density
is given by d = x2 + y 2 = r2 . All together then, the mass of E is
Z
ZZZ
mass(E) =
2π
Z
3
√
Z
d dV =
E
0
0
18−r2
r3 dz dr dθ.
r
4. After a short computation, you find that div(∇f ) = −z(x2 + y 2 ) sin(xy) .
1
5. Since dx = 3 dt and dy = 4t dt, we see that
Z
Z 1
Z 1
2
2
2
3y dx − x dy =
(3(2t ) · 3 − 9t (4t)) dt =
(18t2 − 36t3 ) dt = −3 .
C
0
0
6. From the hint we look for a potential function f (x, y, z) for F, and we see that we can
take f (x, y, z) = x + y + z + xyz so that ∇f = F. Since the vector field is conservative,
we then need to calculate f at the endpoints of C and subtract. When t = 0, we see
that the initial point of C is (0, 0, 0), and when t = π, the final point of C is (π, 0, πe).
Thus
Z
F · dr = f (π, 0, πe) − f (0, 0, 0) = π + πe .
C
7. We use the Divergence Theorem. First we see that div(F) = 4. If we let E denote
the solid region surrounded by S and the unit disk D in the xy-plane, then by the
Divergence Theorem
ZZ
ZZZ
ZZ
F · n dS =
div(F) dV −
F · n dS
S
E
D
ZZ
ZZZ
hx, y, 2zi · h0, 0, −1i dS.
4 dV −
=
D
E
RRR
The triple integral evaluates to 4 times the volume of E: check that
4 dV = 2π.
E
The surface integral over the disk D is 0, because at every point in D we have z = 0,
so the dot product in the integrand is always 0 on D. Thus the answer is 2π .
q 2
2
dx 2
8. (a) Using that ds =
+ dy
dt, we see
+ dz
dt
dt
dt
Z
2
2
2
Z
(x + y + z ) ds =
C
2π
(16 cos2 t + 16 sin2 t + 9t2 )
p
16 sin2 t + 16 cos2 t + 9 dt.
0
Evaluating the integral on the right, we obtain an answer of 160π + 120π 3 .
(b) Let R be the triangular region bounded by the points (0, 0), (2, 0), and (2, 4) in
the xy-plane. Then Green’s Theorem says that
Z
ZZ
2
x2
x
3 2
3x − 3ex dA,
3ye + 2e dx + 2 x + sin y =
C
Z 2RZ 2x
2
=
3x − 3ex dy dx.
0
0
Evaluating the iterated integral, we obtain an answer of 19 − 3e4 .
Z
1
9. (a)
−1
√
Z
1−x2
√
− 1−x2
−4 dy dx
2
(b) Use the usual parametrization of C : x = cos t, y = sin t, 0 ≤ t ≤ 2π. So
dx = − sin t dt, dy = cos t dt. Then
I
Z
Z 2π
F · dr = (2y dx − 2x dy) =
(−2 sin2 t − 2 cos2 t) dt.
C
0
This integral evaluates to −4π .
(c) The integral in (a) is simply −4 times the area of the region of integration, which
is a disk of radius 1. Thus,
Z
1
−1
√
Z
1−x2
√
− 1−x2
−4 dy dx = −4π
as expected from (b).
10. By Stokes’ Theorem, the surface integral in question is the same as a line integral along
the boundary of S. More to the point, if we let C circle of radius 3 in the xy-plane
defined by x = 3 cos t, y = 3 sin t, z = 0, 0 ≤ t ≤ 2π, then
ZZ
I
Z 2π
(curl F) · n dS =
F · dr =
(−9 sin2 t − 9 cos2 t) dt = −18π .
S
C
0
11. In this problem we use Stokes’ Theorem to evaluate the desired line integral in terms
of a surface integral. Given that F = hy, −x, yzi, we calculate that
curl F = hz, 0, −2i.
Now if we let n be the upward unit normal vector on S, then
I
ZZ
(curl F) · n dS.
F · dr =
S
When calculating surface integrals on surfaces defined by the graph of a function, in
this case z = x2 + y 2 , we have some relatively nice formulas. For example, see formula
(8) on p. 922 of Stewart. Then if we let D be the unit disk in the xy-plane,
ZZ
ZZ
ZZ
(curl F) · n dS =
hz, 0, −2i · n dS =
(−z(2x) − 0 · (2y) − 2) dA.
S
S
D
The double integral on the right then evaluates to −2π .
3