Setting the limits of an integral

[docnet]

Homework Statement

Find the area of the unit sphere in R3 enclosed in the offset cone (x-1)^2 +y^2 = z^2

Relevant Equations

x^2 + y^2 + z^2 = 1

(x-1)^2 +y^2 = z^2

Problem: The sphere is parametrized in cylindrical coordinates by:

x = r cosθ
y = r sinθ
z = (1-r^2)^1/2

and intersected by the cone (x-1)^2 +y^2 = z^2.

find the area of the sphere enclosed by the cone using the equation:

da = r/(1-r^2) dr dθ
Attempt at solution:

from the equations for the sphere and cone: r = cosθ describes the intersection in the relevant coordinates.

the values of r ranges from 0 to 1, and θ from -pi/2 to pi/2.

How does one set the limits of integration for the area integral using da = r/(1-r^2) dr dθ ?

I tried setting r from 0 to cosθ and kept getting pi as the area, which is too large.

 

[docnet]

We naturally imagine a map where the area enclosed by the graph of r = cosΘ in polar coordinates is ambiently popped up (projected) onto the surface of the sphere. The induced metric in polar coordinates gives da = r dr dθ. which means we take away the r in our da above and replace it with a 1. This is because the area is multiplied by a factor of 1/(1-r^2)^.5 under this map from polar to cylindrical coordinates, so the factor of expansion approaches infinity as r goes to 1, and 1 as r goes to 0. This makes sense because on the sphere , the slope of the the surface is infinite at the equator and 0 at the pole. so the amount of "stretching" varies accordingly.

integrating 1/(1-r^2)^.5 using the fore-mentioned limits gives pi^2 / 8 as the area of the surface bounded by the intersection with the cone.

 

[LCKurtz]

I tried setting r from 0 to cosθ and kept getting pi as the area, which is too large.

$r$ going from $0$ to $\cos\theta$ is correct. I get $\pi - 2$ for the answer.

 

[docnet]

$r$ going from $0$ to $\cos\theta$ is correct. I get $\pi - 2$ for the answer.

Oops, I am sorry. I made a typo in the original post and did not realize it until now. It should be r/sqrt(1-r^2). We obtained it by the following method.

In the previous part of the problem, we computed the expansion factor r/sqrt(1-r^2) by taking the square root of the determinant of the metric induced by the Euclian metric dx^2 + dy^2 + dz^2 in cylindrical coordinates by this map.

x = r cosθ
y = r sinθ
z = (1-r^2)^1/2

Computing it requires taking the partial derivatives of x, y, z and expanding it out. It simplifies to

(rdr)^2 + 1/(1-r^2) dθ^2

so the matrix of the metric is

r^2 0
0 1/(1-r^2)

Whose square root of the determinant is r/sqrt(1-r^2).Computing ∫∫ r/sqrt(1-r^2) dr dθ gives π which is too large. Solving this problem requires us to realize the expansion factor r/sqrt(1-r^2) needs to be changed. We have to divide this expansion factor by the other expansion factor corresponding to the metric induced by the Euclidian metric on polar coordinates.

r dr dθ

so we have

∫∫ 1/sqrt(1-r^2) dr dθ = pi^2 / 8
with
r: [0, cosθ]
θ : [0, 2pi]

which is the surface area of the sphere enclosed within r = cosθ in cylindrical coordinates.