Triple Integral Problem in Cylindrical Coordinates

[James Brady]

Homework Statement

Use cylindrical coordinates to find the volume of the solid that the cylinder
r = 3cos/theta cuts out of the sphere of radius 3 centered at the origin.

Homework Equations

Why do we evaluate theta from 0 to pi instead of from 0 to 2pi? Don't we want to go all the way around the circle?

The Attempt at a Solution

x^{2} + y^{2} + z^{2} = 9

so z = \sqrt{9 - x^2 + y^2}

z = \sqrt{9 - r^2}

2\int^{2pi}_{0}\int^{3cos(Θ)}_{0}\int^{sqrt(9 - r^2)}_{0}rdzdrdΘ

Instead of integrating from the bottom on the sphere to the top, I did from the x-y plane to the top and multiplied by two to take advantage of symmetry.

\int^{2pi}_{0}\int^{3cos(Θ)}_{0}2rsqrt(9 - r^2)drdΘ

\int^{2pi}_{0}-(\frac{2}{3})(9 - r^2)^{3/2}|^{3cos(Θ)}_{0}dΘ

\int^{2pi}_{0}-(\frac{2}{3})(9 - 9cos^{2})^{3/2} - \frac{-2}{3}9^{3/2}dΘ

The integral goes on and eventually reduces to...

18Θ + 18cosΘ + \frac{1}{3}cos^{3}Θ|^{2pi}_{0}

This gives a value of 113, which is double the correct answer. I'm assuming this is because I was suppose to integrate from 0 to pi instead of from 0 to 2pi, but I don't know why. Aren't we suppose to go around the whole circle with this thing?

 

[LCKurtz]

Draw a polar coordinate graph of $r = 3\cos\theta$. Do $\theta$ from $-\pi/2$ to $\pi/2$ and see what you think.

 

[James Brady]

Ahh... I see it. Thanks.