Question: I need to calculate the volume of solid enclosed by the surface $(x^2+y^2+z^2)^2=x$, using only spherical coordinates.

My attempt: by changing coordinates to spherical: $x=r\sin\phi\cos\theta~,~y=r\sin\phi\sin\theta~,~z=r\cos\phi$ we obtain the Jacobian $J=r^2\sin\phi$. When $\phi$ and $\theta$ are fixed, $r$ varies from $0$ to $\sqrt[3]{\sin\phi\cos\theta}$ (because $r^4=r\sin\phi\cos\theta$). Keeping $\theta$ fixed, we let $\phi$ vary from $0$ to $\pi$. Thus the volume equals: 

$$V=\int\limits\_{0}^{\pi}\int\limits\_{0}^{\pi}\int\limits\_{0}^{\sqrt[3]{\sin\phi\cos\theta}}r^2\sin\phi ~dr ~d\phi ~d\theta=0$$
Which is obviously wrong. What am I doing wrong?

Answer: 
We may as well consider the surface $(x^2+y^2+z^2)=z$ which is a rotational surface $S$ around the $z$-axis. Put $x^2+y^2=:\rho^2$. Then $S$ is given by the equation
$$(\rho^2+z^2)^2=z\geq0\ .$$ It follows that
$$\rho^2(z)=\sqrt{z}-z^2\qquad(0\leq z\leq1)\ ,$$
so that
$$V=\pi\int\_0^1\rho^2(z)\>dz=\pi\>\left({2\over3}z^{3/2}-{1\over3}z^3\right)\biggr|\_0^1={\pi\over3}\ .$$
Byron Schmuland's suggestion is the easiest way to find this volume, 

but using spherical coordinates and symmetry to find the volume bounded by 
$(x^2+y^2+z^2)^2=z\;\;$ 

gives $\;\;\displaystyle V=\int\_0^{2\pi}\int\_0^{\frac{\pi}{2}}\int\_0^{\sqrt[3]{\cos\phi}}\rho^2\sin\phi\;d\rho d\phi d\theta=\frac{\pi}{3}$.