Originally Posted by

**NonCommAlg** ok, in my first post in this thread i tried to simplify things by eliminating z axis but if you want to see things in 3 dimensional here is how it works:

there's no $\displaystyle \theta$ in the equation $\displaystyle \rho=a \sin \phi,$ which means we only need to find the cross section of the surface with some $\displaystyle \theta$ and then rotate the cross section about $\displaystyle z$ axis because $\displaystyle 0 \leq \theta \leq 2\pi.$

so, for example, let $\displaystyle \theta=0.$ then you get $\displaystyle x=\rho \sin \phi=a\sin^2 \phi, \ y = 0, \ z=\rho \cos \phi.$ so the cross section is the circle $\displaystyle x^2+z^2=\rho^2=a^2\sin^2 \phi=ax, \ 0 \leq \phi \leq \pi,$ in the plane $\displaystyle y=0,$ which is

xz plane. sketch this circle and then rotate it about $\displaystyle z$ axis. what do you get? a torus of course!

to answer your second question, $\displaystyle \rho^2 \sin \phi$ is the Jacobian of change of variables from cartisian to spherical coordinates, exactly like $\displaystyle r$ in polar. you always have to put it when calculating the

volume in spherical coordinates. also the tripple integral is over __the region__ bounded by the surface of the torus and not over the surface of the torus. so $\displaystyle 0 \leq \rho \leq a \sin \phi$ and __not__ $\displaystyle \rho=a\sin \phi.$