Third Practice Test Solutions
MATH 224, Fall 2007
Z 1Z
1. Calculate
√
0
Z 1Z
0
1
√
y
1
y
2
yex
dx dy by reversing the order of integration.
x3
Z 1 " 2 x2 #y=x2
2
yex
y e
dy dx =
dx
3
x
2x3
0 0
0
y=0
x=1
Z 1 4 x2
Z 1
xe
x x2
1 x2
e−1
=
e
dx
=
e
dx
=
=
2x3
4
4
0
0 2
x=0
2
yex
dx dy =
x3
Z 1Z
x2
2. Consider a spherical shell E between the spheres x2 + y 2 + z 2 = 1 and
x2 + y 2 + z 2 = 4 with mass density equal to the distance to (0, 0, 0). Find
the total mass.
In spherical coordinates this shell is described by 1 ≤ ρ ≤ 2, 0 ≤ φ ≤ π, and
0 ≤ θ ≤ 2π. The density is given by ρ, so
Z 2Z πZ 2π
Z 2
Z π
2
3
m=
ρ · ρ sin φ dθ dφ dρ = 2π
ρ dρ
sin φ dφ = 15π.
1
0
0
Z
3. Calculate
1
0
y 3 ds, where C is the part of the graph y = 2x3 from (0, 0)
C
to (1, 2).
Standard parameterization is r(t) = ht, 2t3 i, 0 ≤ t ≤ 1, with r0 (t) = h1, 6t2 i,
and thus
Z
Z 1 √
3
y ds =
8t9 1 + 36t4 dt.
C
0
R
Due to a typo (it was supposed to be C y ds), this integral turned out quite
a bit harder to solve than anticipated. Don’t worry if you gave up at this
point. For completeness’ sake, the rest of the solution follows. (Feel free to
ignore it, it is indeed scary.)
Substitute u = 1 + 36t4 to get du = 144t3 dt, and since t6 =
Z
1
37
8
1
(u − 1)3/2 u1/2 du =
144 · 216
3888
Z
(u−1)3/2
216
we get
37
p
(u − 1) u(u − 1)du
1
s
2
Z 37
1
1
1
(u − 1)
− du.
=
u−
3888 1
2
4
Now we substitute r = u −
1
3888
1
2
and get
r
1
1
r−
r2 − dr
2
4
1/2
Z 73/2 r
Z 73/2 r
1
1
1
1
r2 − dr.
=
r r2 − dr −
3888 1/2
4
7776 1/2
4
Z
73/2
The first integral is easy to solve directly (or by substituting v = r2 −1/4), and
the second one can be solved mysteriously by substituting r = (ew + e−w )/4.
Fortunately it is also one of the standard integrals in the back of the textbook
(table of integrals, number 39, with a = 1/2), so we get
" 1
1 2
r −
3888 3
√
37 37
=
54
1
4
3/2 #73/2
1/2
" r
r
1
1
r
1
−
r2 − − ln r + r2 −
7776 2
4 8 √
√
73 37
1
−
+
ln(73 + 12 37) ≈ 4.082
5184
62208
Certainly no integral like this will be on the test.
#73/2
1 4
1/2
4. Which of the following vector fields
Z are conservative? Find a potential for
one of them and use it to calculate
F· dr where C is the arc of the unit
C
circle from (1, 0) to (0, 1) in counterclockwise direction.
F1 (x, y) = hx2 , x2 i
F3 (x, y) = hey , ex i
F2 (x, y) = h2xy, x2 i
F4 (x, y) = hex , ey i
We have to test whether ∂P
= ∂Q
in all four cases. This test shows that
∂y
∂x
F2 and F4 are conservative, whereas F1 and F3 are not. Potentials for the
conservative fields are f2 (x, y) = x2 y and f4 (x, y) = ex + ey . From the
fundamental theorem for line integrals we get that
Z
F2 · dr = f2 (0, 1) − f2 (1, 0) = 02 · 1 − 12 · 0 = 0
ZC
F4 · dr = f4 (0, 1) − f4 (1, 0) = e0 + e1 − (e1 + e0 ) = 0
C
5. Use Green’s Theorem to evaluate
Z √
1 + x2 dx + x(1 + sin y) dy, where C
C
is the unit circle, parameterized in counterclockwise direction. (Don’t even
try to solve this integral directly.)
p
With P (x, y) = (1 + x2 ) and Q(x, y) = x(1 + sin y), we have ∂P
(x, y) = 0
∂y
∂Q
and ∂x = 1 + sin y. Applying Green’s Theorem gives
Z
ZZ
2
sin(1 + x )dx + x(1 + y) dy =
(1 + sin y) dA
C
D
ZZ
ZZ
sin y dA = π + 0 = π,
1 dA +
=
D
D
where the first integral is just the area of the unit disk D, and the second one
is zero by symmetry. If this is not immediately clear, here is one example
where it might be easier not to use polar coordinates. The second integral is
Z 1Z √1−x2
Z 1
√
y= 1−x2
√
sin
y
dy
dx
=
[−
cos
y]
dx
√
y=− 1−x2
−1 − 1−x2
−1
Z 1
√
√
2
2
=
− cos(− 1 − x ) + cos 1 − x dx = 0,
−1
√
√
since cos(− 1 − x2 ) = cos 1 − x2 .