Triple Integral with spherical coordinates Find the volume of the region bounded by $(x^2 + y^2 + z^2)^2 = x$.

I used spherical coordinates to get $p^3 = \sin \phi \cos \theta$ so does this imply that

$ 0&lt;= p &lt;= (\sin \phi \cos \theta)^{1/3}$. And if so would there be no restrictions to 

$\phi $ and $\theta$ other than the regular ($ 0 \leq \phi \leq \pi$ and $ 0 \leq \theta \leq 2\pi$)? I tried this but the maple returns $0$ when it evaluates the triple integral.
The volume is $V = \pi/3$. First observe that, if $(x,y,z)$ belongs to the region, then $0 \leq x \leq 1$. Fix now $0 \leq x_0 \leq 1$. The points on the plane $\{x = x_0\}$ form a circle of radius $r(x_0)$, where
$$
r(x_0)^4+2r(x_0)^2x_0^2+x_0^4 = x_0,
$$
i.e.,
$$
r(x_0)^2 = \sqrt{x_0} - x_0^2.
$$ 

Now we integrate by sections to get
$$
V = \int_0^1 \pi(\sqrt{x}-x^2) dx = \frac{\pi}{3}.
$$
This is the same as the volume of the region bounded by $$(x^2+y^2+z^2)^2=z$$  You may find this easier to evaluate in spherical coordinates.