SOLUTION OF HW5
MINGFENG ZHAO
October 23, 2011
1. [Page 94, Problem 15] The graphs of f (x) = x2 and g(x) = cx3 , where c > 0, intersect at the points
(0, 0) and
1 1
c , c2
. Find c so that the region which lies between these graphs and over the interval
0, 1c has area 23 .
Proof. In fact, we have
2
3
1
c
Z
[f (x) − g(x)] dx
=
0
1
c
Z
=
[x2 − cx3 ] dx
0
=
1
x3
cx4 c
−
3
4 0
=
1
c
− 4
3
3c
4c
=
1
1
− 3
3c3
4c
=
1
.
12c3
1
2
MINGFENG ZHAO
That is, we have c3 =
1
12× 23
= 18 , hence we get
c=
1
.
2
2. [Page 94, Problem 19] Let C denote the unit circle, whose Cartesian equation is x2 + y 2 = 1. Let
E be the set of points obtained by multiplying the x-coordinate of each pint (x, y) on C be a constant
factor a > 0 and the y-coordinate by a constant factor b > 0. The set E is called an ellipse. (When
a = b, the ellipse is another circle. )
a. Show that each point (x, y) on E satisfies the Cartesian equation
x 2
a
+
y 2
b
= 1.
b. Use properties of the integral to prove that the region enclosed by this ellipse is measurable
and this its area is πab.
Proof. a. For any point (x, y) ∈ E, then
x y
a, b
x 2
a
∈ C, that is,
+
y 2
b
= 1.
SOLUTION OF HW5
3
b. Let D be the region enclosed by this ellipse E. For any (x, y) ∈ E, by the result of the part a, we
have
x 2
a
+
y 2
b
= 1.
Then we get
b2
x2
y = b · 1 − 2 = b2 − 2 · x 2 .
a
a
2
2
Then
Z
Area (D)
=
dxdy
D
s
a
x2
b2 · 1 − 2 dx
a
−a
r
Z a
x2
=
2b · 1 − 2 dx
a
−a
Z ar
x2
1 − 2 dx
= 2b ·
a
−a
Z 1r
a2 y 2
1 − 2 a dy Let x = ay
= 2b ·
a
−1
Z 1p
= 2ab ·
1 − y 2 dy.
Z
=
2
−1
On the other hand, for the unit disc, we have
Z
π
1
p
2 1 − x2 dx
=
−1
1
Z
=
p
1 − y 2 dy
2
Let x = y.
−1
Hence, we get
Area (D) = πab.
3. [Page 105, Problem 23] Evaluate the integral
Z
0
π
1
+ cos t dt.
2
4
MINGFENG ZHAO
Proof. Since we know that cos t ≥ − 21 when 0 ≤ t ≤
Z
0
π
1
+ cos t dt
2
2π
2
Z
=
0
2π
2
Z
0
=
and − 12 > cos t when
Z π
1
+ cos t dt +
2
2π
2
=
2π
3 ,
1
+ cos t
2
< t ≤ π, so we have
1
+ cos t dt
2
π
Z
2π
3
dt −
2π
2
1
+ cos t
2
dt
2π
π
3
t
t
+ sin t −
+ sin t 2
2
2π
0
3
1 2π
2π 1 π
2π
×
+ sin
− · + sin
2
3
3
2 3
3
√
√
3 π
3
π
+
− +
3
2
6
2
=
=
√
=
3+
π
.
6
4. [Page 111, Problem 13] Sketch the graph and compute the area of radial set offover the Four-leaf
colorer:
f (θ) =
p
| cos 2θ|,
0 ≤ θ ≤ 2π.
Proof. By the Theorem 2.6, in Page 110, we know that
Area
=
=
Z 2π
1
[f (θ)]2 dθ
·
2 0
Z
1 2π p
[ | cos 2θ|]2 dθ
2 0
SOLUTION OF HW5
=
=
=
=
1
2
Z
1
2
Z
1
4
Z
5
2π
| cos(2θ)| dθ
0
4π
| cos x| ·
0
1
dx
2
Let θ =
x
2
4π
| cos x| dx
0
2.
5. [Page 114, Problem 13] What volume of material is removed from a solid sphere of radius 2r by
drilling a hole of radius r through the center?
6
MINGFENG ZHAO
Proof. By the Theorem 2.7, in Page 113, we have
Z
V
=
2r
ar (u)
−2r
Z −√3r
du
√
2
Z
2
π[(2r) − u ] du +
=
−2r
√
− 3r
Z
=
π[4r2 − u2 ] du +
Z
−2r
=
=
=
3r
√
πr du +
− 3r
√
3r
√
Z
2
− 3r
πr2 du +
Z
2r
√
π[(2r)2 − u2 ] du
3r
2r
√
π[4r2 − u2 ] du
3r
−√3r
2r
√3r
u3 u3 2
2 √
2
π · 4r u −
+
πr
u
+
π
·
4r
u
−
− 3r
3 −2r
3 √3r
√
√ 3 √ 3
√ 3 √ 3
8r3
8r3
3
3
3
+ 2 3πr + π · 8r −
− 4 3r + 3r
π · −4 3r + 3r + 8r −
3
3
√
32
− 4 3 πr3 .
3
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu