# Transformation to spherical coordinate system

If I have a sphere $T: x^{2}+y^{2}+z^{2}\leqslant 10z$ by transformation to the spherical coordinate system by the: $x=r\cos\theta\sin\varphi\\ y=r\sin\theta\sin\varphi\\ z=r\cos\varphi$

What is values for $\varphi$ I will get?

$0\leqslant\varphi\leqslant\pi$ or $0\leqslant\varphi\leqslant\pi/2$ and why??

Thanks!

• Something looks odd, the coordinates for $r$ a constant is a spherical shell (surface) centered at the origin, but $T$ is not centered at the origin. Have you read the problem correctly? – Brady Trainor Jun 2 '13 at 2:09

$x^2+y^2+z^2 \leq 10z$ is the same as $x^2+y^2+(z-5)^2\leq 25$ so we are dealing with a solid ball of radius 5 centered at $(0,0,5)$.
Naively we can translate this inequality to $\rho^2 \leq 10\rho \cos(\phi)$ so that $\rho \leq 10\cos(\phi)$.
I have drawn a ray emanating from the origin out to the sphere [whose equation is $\rho=10\cos(\phi)$]. The angle $\phi$ should sweep from the $z$-axis down to $\phi=\pi/2$. Notice that at this point $\rho=10\cos(\pi/2)=0$ (so we should stop).
Notice that if you had allowed $\phi$ to continue to go past $\pi/2$, cosine and thus $\rho$ would become negative (which indicates something isn't quite right).
$0 \leq \rho \leq 10\cos(\phi)$, $0 \leq \phi \leq \pi/2$, and $0 \leq \theta \leq 2\pi$.