Setting up a triple integral in cylindrical coordinates?

[VinnyCee]

The problem says to find the volume of material cut from the solid sphere,

r^2 + z^2 \le 9

by the cylinder,

r = 3\sin\theta

I don't know how to graph the first equation, but I can do the second in polar coordinates. How do I go about converting to use cylindrical coordinates?

 

[SpaceTiger]

I don't know how to graph the first equation

It's a sphere of radius 3 centered at the origin:

r^2+z^2=x^2+y^2+z^2=9

but I can do the second in polar coordinates. How do I go about converting to use cylindrical coordinates?

It's already in cylindrical coordinates. The equation is implying that r=3sin(\theta) for all z.

 

[VinnyCee]

I did a crude sketch and came up with this integral

Is this correct limits for the problem?

\int_{0}^{2\pi}\int_{0}^{3\sin\theta}\int_{-\sqrt{9 - r^2}}^{\sqrt{9 - r^2}}\;dz\;r\;dr\;d\theta

 

[HallsofIvy]

"Cylindrical coordinates" is simply polar coordinates with "z" added.
r= 3 sin \theta, a circle with center at (0, 3/2), radius 3/2, in polar coordinates is a cylinder running parallel to the z axis in cylindrical coordinates.
Since r²= x²+ y², the sphere, x²+ y²+ z²= 9 is r²+ z²= 9 in cylindrical coordinates.

 

[SpaceTiger]

Looks good to me.

 

[VinnyCee]

Many Thanks

Thank you both. When I have more time on my hands, I will be sure and return the favor to someone here someday