How to Find the Volume of a Region Enclosed by \(\rho = 2 \sin\theta\)?

[VinnyCee]

Here is the problem:

Find the volume of the region enclosed by the spherical coordinate surface $$\rho = 2 \sin\theta$$, using spherical coodinates for the limits of the integral.

Here is what I have:

I don't know if this is right, but here it is $$\int_{0}^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\sin\theta}\;\rho^2\;\sin\theta\;d\rho\;d\phi\;d\theta$$

 

[dextercioby]

It looks okay.Did u plot it?Heh,u we use $r,\varphi,\vartheta$ as the spherical coordinates...

Daniel.

 

[VinnyCee]

Whoops!

I posted the wrong thing. Instead of $$2\sin\theta$$ for the first integrals upper limit, it should be $$2\sin\phi$$?

$$\int_{0}^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\sin\phi}\;\rho^2\;\sin\theta\;d\rho\;d\phi\;d\theta$$

So confusing!

Is that wrong now?

The graph looks like a donut without the hole centered at the origin. I will post it shortly.

 

[dextercioby]

No,it looks okay.

Daniel.