SOLUTION OF HW6
MINGFENG ZHAO
October 23, 2011
1. [Page 45, Problem 10] If x > 1, prove by induction that xn > x for every integer n ≥ 2. If
0 < x < 1, prove that xn < x for every integer n ≥ 2.
Proof. Claim I: If x > 1, then xn > x for all n ≥ 2.
When n = 2, we know that
x2 − x = x(x − 1) > 0.
So x2 > x, which implies that the Claim I is true for n = 2. Assume the Claim I is true for
n = k ≥ 2, that is,
xk > x.
Now look at n = k + 1, we have
xk+1 − x
= x(xk − 1)
>
x(x − 1)
> 0.
So xk+1 > x, which implies that the Claim I is true for n = k + 1. Then by the induction, we can
conclude that xn > x for all n ≥ 2.
Claim II: If 0 < x < 1, then xn < x for all n ≥ 2.
When n = 2, we know that
x2 − x = x(x − 1) < 0.
1
2
MINGFENG ZHAO
So x2 < x, which implies that the Claim II is true for n = 2. Assume the Claim II is true for
n = k ≥ 2, that is,
xk < x.
Now look at n = k + 1, we have
xk+1 − x
= x(xk − 1)
x(x − 1)
<
< 0.
So xk+1 < x, which implies that the Claim II is true for n = k + 1. Then by the induction, we can
conclude that xn < x for all n ≥ 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 HW6
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
x2
b2
y 2 = b2 · 1 − 2 = b2 − 2 · x 2 .
a
a
Then
Z
Area (D)
=
dxdy
D
a
s
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
r
Z 1
a2 y 2
= 2b ·
1 − 2 a dy Let x = ay
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
=
−1
p
2 1 − x2 dx
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MINGFENG ZHAO
Z
=
1
2
p
1 − y 2 dy
Let x = y.
−1
Hence, we get
Area (D) = πab.
3. [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. The graph should be:
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
Z
1 2π
| cos(2θ)| dθ
2 0
Z
1 4π
1
x
| cos x| · dx Let θ =
2 0
2
2
Z
1 4π
| cos x| dx
4 0
SOLUTION OF HW6
=
5
1.
4. [Page 114, Problem 15] A solid has a circular base of radius 2. Each cross section cut by a plane
perpendicular to a fixed diameter is an equilateral triangle. Compute the volume of the solid.
Proof. Recall that if the equilateral triangle’s side length is 2d, then the area should be
√
3d2 .
By the Theorem 2.7, in Page 113, we have
Z
V
2
=
a2 (u) du
−2
Z
2
=
√
3 · [22 − u2 ] du
−2
=
√ Z
3
2
[4 − u2 ] du
−2
2
u3 =
3 · 4u −
3 −2
√
16
3 · 16 −
=
3
√
32 3
=
.
3
√
6
MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu