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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?

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up vote 2 down vote accepted

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: $$ {0}\leqslant{\theta}\leqslant{2\pi}, \\ {0} \leqslant \varphi \leqslant \dfrac{\pi}{2}, \\ {0}\leqslant{\rho}\leqslant 2\cos{\varphi} \, \sin^2{\varphi}. $$

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In spherical, $\varphi \in [0, \pi]$. Source: – apnorton Dec 15 '12 at 0:28
Thanks, I edit my answer. – M. Strochyk Dec 15 '12 at 0:29
Umm... the range on $\varphi$ needs to go from $0$ to $\pi$ – apnorton Dec 15 '12 at 0:36
@anorton From the surface equation $(x^2 + y^2 + z^2)^2 = 2z(x^2 + y^2)$ we can conclude that $z$ must be nonnegative, which is possible only if ${0}\leqslant {\varphi}\leqslant{\dfrac{\pi}{2}}$. – M. Strochyk Dec 15 '12 at 0:46
Oh. Duh. Nevermind. – apnorton Dec 15 '12 at 0:54

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