Set up triple integral's boundary for $x^{2} + (y-a)^{2} + z^{2}=a^{2}$ in spherical coordinates.

Question

I have trouble with setting up triple integral's boundary for $\rho$. Solid object's equation is $x^{2} + (y-a)^{2} + z^{2}=a^{2}$,which is a sphere centered at (0,a,0), in spherical coordinates. Note: a is just a constant.

Answer 1

If you do not want to translate the sphere so that its center lies in the origin, you will have to plug the spherical coordinate change in the inequality $$\tag{*} x^2+(y-a)^2+z^2 \le a^2.$$ Namely, using the following convention: $$ \begin{cases} x=r\sin \theta \cos \phi \\ y=r\sin \theta \sin \phi \\ z=r\cos \theta \end{cases}, \qquad r\in [0, \infty),\ \theta\in [0, \pi],\ \phi\in [0, 2\pi), $$ one has $$ \tag{*} r^2 -2ar\sin\theta\sin\phi \le 0.$$ Note that this inequality has no solutions for $\phi\notin[0, \pi]$. (This is easy to visualize geometrically as well).

This means that the solid ball can be rewritten in spherical coordinates as follows: $$ \left\{ (r, \theta, \phi)\ :\ r\le 2a\sin\theta\sin\phi, \phi\in[0, \pi],\ \theta\in[0, \pi] \right\}. $$