2. hw3 Partial Solution
(18) Since the volume of a three dimensional object is invariant under translation, we may assume that a = b = c = 0. The intersection of x = t plane with S has area A(t) where
A(t) =
πBC
1 − t2
A2
if −A ≤ t ≤ A
0 otherwise.
The volume of S is given by Vol (S) =
Z A
−A
A(t)dt = πBC Z A
−A
1 − t2
A2
dt = 4
3πABC.
The boundary of an ellipsoid in general is not a surface of revolution. Hence we need to make some assumption to obtain a surface of revolution. Let us assume B = C = R. Then the surface ∂S is obtained from revolving the curve
x2 A2 + y2
R2 = 1, y ≥ 0, about the x-axis. Let us consider the parametrization
x = A cos θ, y = R sin θ, 0 ≤ θ ≤ π.
Thus the surface area is given by A(∂S) = 2π
Z π 0
R sin θp
A2sin2t + R2cos2θdt
= 2πR Z π
0
sin θp
A2+ (R2− A2) cos2θdt.
Let u = cos θ. Then
A(∂S) = 2πR Z 1
−1
pA2+ (A2− R2)u2du.
The integral can be expressed in terms of elementary function for any A > 0.
(19) In this exercise, we only need to assume that Ω is “simply connected.” The notion “simply connectedness” is a “topological property” which is not introduced yet. Pictorially, it is a region with no holes inside. Mathematically, the term simply connected is not easy defined.
Therefore, we assume that Ω is convex; this notion is much easier understood.
Assume that ∂Ω is positively oriented and parametrized by smooth functions x = x(t), y = y(t) for a ≤ t ≤ b. The area of Ω is given by
A(Ω) = 1 2
Z b a
(x(t)y0(t) − x0(t)y(t))dt.
By Cauchy-Schwarz inequality, we know x(t)y0(t) − x0(t)y(t) ≤p
x(t)2+ y(t)2p
x0(t)2+ y0(t)2≤ Rp
x0(t)2+ y0(t)2. Thus
A(Ω) ≤ R 2
Z b a
px0(t)2+ y0(t)2dt = Rl 2 .
(20) Using the area formula, A(Ω1) = A(Ω2). For (b), you can use elementary geometry to prove it or using the notion of convexity/concavity of a function. Let F (x) = √
1 + x2, x ∈ R.
Then F0(x) = x/√
1 + x2and F00(x) =
√1 + x2−√x2
1+x2
1 + x2 = 1
(1 + x2)3/2 > 0.
1
2
Thus the function F is concave up. It follows from the arclength formula that l(C1) =
Z 1 0
p1 + f0(x)2dx + Z 1
0
p1 + g0(x)2dx
l(C2) = 2 Z 1
0
s
1 + f0(x) + g0(x) 2
2 dx.
Using (b), we find l(C2) ≤ l(C1).
