Find the volume bounded by the surface $(x^2 + y^2 + z^2)^2 = 2z(x^2 + y^2)$
I have $x = \rho \sin\phi \cos\theta$, $y = \rho \sin\phi \sin\theta$, $z = \rho \cos\phi$.
Therefore, $(x^2 + y^2 + z^2)^2 = 2z(x^2 + y^2)$ can be written as
$$(\rho^2)^2 = 2(\rho \cos\phi) [(\rho \sin\phi \cos\theta)^2 + (\rho \sin\phi \sin\theta)^2]\Rightarrow \rho = 2\cos\phi \sin^2\phi$$
so using the change of co-ordinates, I have the integral
$$\iiint \rho^2 \sin\phi d\rho d\phi d\theta$$
Now i need to find the limits of integration, but I cannot even visualize this surface. What are the limits of integration, and is my work so far correct?
In spherical coordinates ${\varphi}\in \left[0, \; {\pi}\right].$ Because ${\rho}\geqslant{0},$ then $\cos{\varphi \geqslant{0}} \Rightarrow {{0}\leqslant {\varphi}\leqslant{\dfrac{\pi}{2}}}.$ Then limits of integration are: $$ {0}\leqslant{\theta}\leqslant{2\pi}, \\ {0} \leqslant \varphi \leqslant \dfrac{\pi}{2}, \\ {0}\leqslant{\rho}\leqslant 2\cos{\varphi} \, \sin^2{\varphi}. $$