110.202. Calculus III
2004 Summer
Midterm II
7/22/2004 1:00PM
1. Evaluate
Z 1Z
0
y
0
Z
x
√
3
x2
0
x
dzdxdy.
+ z2
[Solution]
x
2
Dividing x2 +z
2 by x , the integrand becomes
z
.
x
So, xdu = dz. Therefore,
Z 1 Z y Z √x
3
0
=
=
=
=
=
=
0
Z 1Z
0
y
Z
x2
x
dzdxdy
+ z2
√1
3
1
dudxdy
1 + u2
0
0
0
¸
Z 1Z y∙
¯ √1
−1 ¯ 3
tan u u=0 dxdy
0
0
Z 1Z y
π
dxdy
0
0 6
Z 1
π
ydy
0 6
à ¯1 !
π y 2 ¯¯
6
2 ¯y=0
π
.
12
¥
2. Evaluate
Z 1Z
0
[Solution]
π
4
tan−1 y
¡ 5 ¢
sec x dxdy.
1
1
x
z 2
x
1+(
)
. Let u =
2
By drawing the graph of the region, we change the order of
integration into
Z π Z tan x
Z 1Z π
4
4
¡ 5 ¢
¡ 5 ¢
sec x dxdy =
sec x dydx
0
tan−1 y
0
0
=
Z
π
4
0
=
Z
0
=
Z
1
π
4
¡ 5 ¢ ¯tan x
sec x y ¯y=0 dx
¡ 5 ¢
sec x tan xdx
√
2
¡ 4 ¢
sec x d sec x
¯√2
sec5 x ¯¯
=
5 ¯sec x=1
¡√ ¢5
2
1
=
−
5
√5
4 2−1
.
=
5
¥
3. Integrate f (x, y, z) = xyz along the path
c (t) = (cos t, sin t, t)
where 0 ≤ t ≤ 2π.
[Solution]
√
We have c0 (t) = (− sin t, cos t, 1). Therefore, kc0 (t)k = 2.
Hence,
Z
Z 2π
f ds =
xyz kc0 (t)k dt
0
c
Z 2π
√
=
(cos t) (sin t) t 2dt.
0
Set u = t and dv = cos t sin tdt. Then, we have du = dt and
v = 12 sin2 t. By integration by parts, we have
µ
¶¯2π
¶
Z 2π µ
Z 2π
1 2
1 2 ¯¯
(cos t) (sin t) tdt = t
sin t ¯ −
sin t dt
2
2
0
0
t=0
¶
Z µ
1 2π 1 − cos 2t
dt
= −
2 0
2
π
= − .
2
3
Therefore,
Z
Z
√
(cos t) (sin t) t 2dt
0
√
π 2
= −
.
2
f ds =
c
2π
¥
4. Let B be the region in the first quadrant bounded by the curves
xy = 1, xy = 3, x2 − y 2 = 1 and x2 − y 2 = 4. Evaluate
ZZ
¡ 2
¢
x + y 2 dxdy
B
using the change of variables u = x2 − y 2 and v = xy.
[Solution]
Since B is bounded by x2 − y 2 = 1 and x2 − y 2 = 4, we have
1 ≤ u ≤ 4. Since B is bounded by xy = 1 and xy = 3, we have
1 ≤ v ≤ 3. Let B ∗ be the region in the uv plane bounded by
1 ≤ u ≤ 4 and 1 ≤ v ≤ 3. Let T (x, y) = (u, v) = (x2 − y 2 , xy) :
B → B ∗ be a transformation. We have the Jacobian of T is
¯
¯
¡ 2
¢
∂ (u, v) ¯¯ 2x −2y ¯¯
2
=
2
x
+
y
=¯
.
y
x ¯
∂ (x, y)
By the change of variables theorem, we have
ZZ
ZZ
£ ¡ 2
¢¤
2
1 · 2 x + y dxdy =
1dudv.
B∗
B
RR
Since B∗ 1dudv is the area of B ∗ , we have B∗ 1dudv =
(4 − 1) (3 − 1) = 6. Hence, we have
ZZ
ZZ
ZZ
£ ¡ 2
¡ 2
¢
¢¤
2
2
1 · 2 x + y dxdy =
1dudv = 6.
x + y dxdy =
2
RR
B
B
This implies that
ZZ
B
¥
5. Evaluate
ZZZ
D
B∗
¡ 2
¢
x + y 2 dxdy = 3.
¡ 2
¢
x + y 2 + z 2 xyzdxdydz
over the sphere D = {(x, y, z) | x2 + y 2 + z 2 ≤ 2}.
[Solution]
4
Using spherical coordinates, let x = ρ sin φ cos θ, y = ρ sin φ sin θ
and z = ρ cos φ where 0 ≤ ρ ≤ 2, 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.
The Jacobian is ρ2 sin φ. Therefore,
ZZZ
=
=
=
=
¡ 2
¢
x + y 2 + z 2 xyzdxdydz
D
Z 2 Z π Z 2π
ρ2 (ρ sin φ cos θ) (ρ sin φ sin θ) (ρ cos φ) ρ2 sin φdθdφdρ
0
0
0
Z 2 Z π Z 2π
¡
¢
ρ7 cos φ sin3 φ (cos θ sin θ) dθdφdρ
¶ µZ 2π
¶
µ0Z 20 0 ¶ µZ π
7
3
ρ dρ ·
cos φ sin φdφ ·
cos θ sin θdθ
0
0
0
à ¯ ! Ã
¯π ! Ã 4 ¯2π !
2
sin4 φ ¯¯
sin θ ¯¯
ρ8 ¯¯
·
·
¯
¯
8 ρ=0
4 φ=0
4 ¯θ=0
28
·0·0
8
= 0.
=
¥
6. Find the center of mass of the region between y = x2 ans y = x
if the density function is
δ (x, y) = x + y.
[Solution]
By the formula of the center of mass, we have
RR
xδ (x, y) dxdy
x̄ = RRD
δ (x, y) dxdy
D
and
RR
yδ (x, y) dxdy
.
ȳ = RRD
δ (x, y) dxdy
D
5
By drawing down the region D, the region D can be described
as 0 ≤ x ≤ 1 and x2 ≤ y ≤ x. Therefore,
Z 1Z x
ZZ
δ (x, y) dxdy =
(x + y) dydx
0
x2
D
¶¯x #
Z 1 "µ
y 2 ¯¯
xy +
dx
=
2 ¯y=x2
0
¶ µ
¶¸
Z 1 ∙µ
x2
x4
2
3
x +
=
− x +
dx
2
2
0
¶¯1
µ 3
x4 x5 ¯¯
x
−
−
=
2
4
10 ¯
x=0
18
,
=
120
ZZ
D
xδ (x, y) dxdy =
Z 1Z
0
x
x (x + y) dydx
x2
¶¯x #
Z 1 "µ
xy 2 ¯¯
2
x y+
dx
=
2 ¯y=x2
0
¶ µ
¶¸
Z 1 ∙µ
x3
x5
3
4
x +
=
− x +
dx
2
2
0
µ 4
¶¯1
3x
x5 x6 ¯¯
=
−
−
8
5
12 ¯
x=0
11
,
=
120
and
Z 1Z x
ZZ
yδ (x, y) dxdy =
y (x + y) dydx
0
x2
D
¶¯x #
Z 1 "µ 2
xy
y 3 ¯¯
dx
=
+
2
3 ¯y=x2
0
¶ µ 5
¶¸
Z 1 ∙µ 3
x
x3
x
x6
=
+
−
+
dx
2
3
2
3
0
µ 4
¶¯1
5x
x6 x7 ¯¯
=
−
−
24
12 21 ¯
x=0
13
.
=
168
6
Hence, we have x̄ =
¥
11
120
18
120
=
11
18
and ȳ =
13
168
18
120
=
65
.
126
7. Determine whether the integral
ZZ
x+y
dxdy
2
2
D x + 2xy + y
exists where D = [0, 1] × [0, 1]. If it exists, compute its value.
[Solution]
1
The integrand can be simplified to x+y
when x + y 6= 0. This
integral is improper whenever x + y = 0. In our region D,
there is only one point (0, 0) which satisfies x + y = 0. Let
1
Dδ = [δ, 1] × [0, 1]. We have Dδ → D as δ → 0 and x+y
is
well-defined and integrable in Dδ . Therefore,
ZZ
x+y
dxdy
2
2
D x + 2xy + y
Z 1Z 1
1
dydx
= lim
δ→0 δ
0 x+y
Z 1h
i
= lim
(ln |x + y|)|1y=0 dx
δ→0 δ
Z 1
[ln (x + 1) − ln (x)] dx
= lim
δ→0 δ
©
ª
= lim [((x + 1) ln (x + 1) − x) − (x ln x − x)]|1x=δ
δ→0
= lim {2 ln 2 − (δ + 1) ln (δ + 1) + δ ln δ} .
δ→0
1
¥
Since lim δ ln δ = lim ln1δ = lim −δ1 = lim −δ = 0, we have
δ→0
δ→0 δ
δ→0 δ2
δ→0
RR
x+y
dxdy
=
2
ln
2.
D x2 +2xy+y2
8. During an interleague game against the Philadelphia Phillies,
Oakland A’s third basemen Eric Chavez was given a day off to
get Mark McLemore’s big bat into the lineup. Overcome by
the boredom of watching the game from the sidelines, Chavez
decides to steal the Phillie Phanatic’s four-wheeler. At time t =
0, Chavez begins to speed away in the four-wheeler, following
the path defined by
and
x (t) = 3 − 5 cos (πt)
y (t) = 1 + 5 sin (πt)
7
where t ≥ 0. How far does Chavez travel during the first 5
seconds of his escape?
[Solution]
The Chavez’s getaway path is c (t) = (x (t) , y (t)) = (3 − 5 cos (πt) , 1 + 5 sin (πt))
for t ≥ 0. The length of the path after 5 seconds is
Z 5
L (c) =
kc0 (t)k dt
0
Z 5q
[−5π sin (πt)]2 + [5π cos (πt)]2 dt
=
0
Z 5
5πdt
=
0
= 25π.
¥