spherical coordinate integral at (0,0,1)

Question

I am trying to set up and solve the integral $\iiint \sqrt(x^2+y^2+z^2)$ using the integration order d$\rho$ d$\phi$ d$\theta$ for the region $x^2+y^2+z^2=2z$. I know how to set up the integral and solve given the center is the origin or even if there is a cone attached to the bottom of the sphere for lower bounds but I can not figure out how to set this up since the center is at (0,0,1). please help and be specific. I do not know very much

Answer 1

$x = \rho \cos \theta \sin \phi\\ y = \rho \sin \theta \sin \phi\\ z = \rho \cos \phi$

$x^2 + y^2 + z^2 = 2z\\ \rho^2 = 2\rho \cos\phi\\ \rho = 2\cos\phi$

Since the sphere is above the xy plane, we do not need to take $\phi$ beyond $\frac \pi 2$

$\int_0^{2\pi}\int_0^{\frac {pi}{2}}\int_0^{2\cos\phi} \rho (\rho^2\sin\phi) d\rho\;d\phi\;d\theta$