Volume enclosed by $(x^2+y^2+z^2)^2=x$ 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?

It's a solid of revolution.  Let $h=\sqrt{y^{2}+z^{2}}$, then $(x^{2}+h^{2})^{2}=x$.
$\therefore \; h^{2}=\sqrt{x}-x^{2} \:$ where $\, 0\leq x \leq 1$.
\begin{align*}
  V &= \int_{0}^{1} \pi h^{2} dx \\
    &= \pi \int_{0}^{1} \left( \sqrt{x}-x^{2} \right) dx \\
    &= \pi \left[ \frac{2}{3} x^{\frac{3}{2}}-
                  \frac{1}{3} x^{3} \right]_{0}^{1} \\
    &= \frac{\pi}{3}
\end{align*}
