Volume of a sphere in cylindrical coordinates

[Thomas Kieffer]

Homework Statement

A sphere of radius 6 has a cylindrical hole of radius 3 drilled into it. What is the volume of the remaining solid.

The Attempt at a Solution

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I am able to solve this using cylindrical coordinates but I'm having trouble when I try to solve it in spherical coordinates. the correct answer is $4\pi\cdot 3^{7/2}$ however I get $4\pi\cdot 3^{3/2}$. The problem is with the bounds of the integration, I checked my working with wolfram.

$$\int_{0}^{2\pi} \int_{-\pi/3}^{\pi/3} \int_{3/sin(\theta)}^{6} r^2sin(\theta)dr d\theta d\phi$$

The first two bounds are obvious. phi ranges the entire circle and theta ranges from the intersects of the edge of the sphere and the drill. The bounds of r I found by converting from Cartesian to polar coordinates. Obviously the upper bound is six.

Converting from cartesian
$$x^2 + y^2 = 9 \qquad r^2sin^2(\theta) (sin^2(\phi) + cos^2(\phi))= 9$$
$$r = 3/sin(\theta)$$

 

[Dick]

Homework Statement

A sphere of radius 6 has a cylindrical hole of radius 3 drilled into it. What is the volume of the remaining solid.

The Attempt at a Solution

[/B]
I am able to solve this using cylindrical coordinates but I'm having trouble when I try to solve it in spherical coordinates. the correct answer is $4\pi\cdot 3^{7/2}$ however I get $4\pi\cdot 3^{3/2}$. The problem is with the bounds of the integration, I checked my working with wolfram.

$$\int_{0}^{2\pi} \int_{-\pi/3}^{\pi/3} \int_{3/sin(\theta)}^{6} r^2sin(\theta)dr d\theta d\phi$$

The first two bounds are obvious. phi ranges the entire circle and theta ranges from the intersects of the edge of the sphere and the drill. The bounds of r I found by converting from Cartesian to polar coordinates. Obviously the upper bound is six.

Converting from cartesian
$$x^2 + y^2 = 9 \qquad r^2sin^2(\theta) (sin^2(\phi) + cos^2(\phi))= 9$$
$$r = 3/sin(\theta)$$

Your $\theta$ limits look wrong. $\theta$ is supposed to be in the range $0$ to $\pi$.

 

[Thomas Kieffer]

Your $\theta$ limits look wrong. $\theta$ is supposed to be in the range $0$ to $\pi$.

Why? That's definitely not correct.

 

[Dick]

Why? That's definitely not correct.

Then you are mixing up different conventions for spherical coordinates. You quoted the volume element as $r^2 \sin(\theta)$. That's negative for $\theta=-\pi/3$. The volume element shouldn't be negative anywhere.

 

[Thomas Kieffer]

The change of coordinates I used was. I don't think I'm mixing up my convention. I also don't understand why the volume element has to be positive.

$x=rcos\phi sin\theta$
$x=rsin\phi sin\theta$
$z=rcos\theta$

If you work through the integral you find
$\int_{0}^{2\pi} [-72 cos\theta]^{\pi/3}_{-pi/3} + [9cot\theta]^{\pi/3}_{-pi/3} d\phi$

And the $-72cos\theta$ must cancel out so the bounds must be a=-b . With the bounds I have I get an answer that is exactly 1/9th the correct answer.

 

[Dick]

I changed the bounds to 0 to pi/3 and multiplied the integral by 2 and now I get the right answer. Thanks. So I need to be careful whenever I'm finding volumes or surface area that the volume element/ surface element is always positive?

No, you don't need to be careful. In your chosen coordinates the volume element should ALWAYS be positive. In my convention, $\theta$ is the angle between between the point and the positive $z$ axis. So it starts at $0$ and goes to $\pi$ at the negative $z$ axis. I changed the $\theta$ range to $\pi/6$ to $5\pi/6$ and got the right answer.

 

[Thomas Kieffer]

No, you don't need to be careful. In your chosen coordinates the volume element should ALWAYS be positive. In my convention, $\theta$ is the angle between between the point and the positive $z$ axis. So it starts at $0$ and goes to $\pi$ at the negative $z$ axis. I changed the $\theta$ range to $\pi/6$ to $5\pi/6$ and got the right answer.

I thought the angle was relative to the xy plane. Thanks for your help.