volume of surface given by $(x^2+y^2+z^2)^2=x$ A question asks me to find the volume of the surface $(x^2+y^2+z^2)^2=x$ this looks like a very difficult triple integral to evaluate using cartesian coordinates, so I though I would describe the set in spherical coordinates. 
Doing so I got $p^2=sin(\phi)cos(\theta)$. But this still seems like a very difficult integral to evaluate. Am I doing something wrong, or am I completely missing something?
 A: You can use polar coordinates to evaluate the integral. However, you should swap
the roles of $x,y,z$ directions to simplify the expressions.
In polar coordinates $(r,\theta,\phi) \mapsto ( r\cos\theta, r\sin\theta\cos\phi, r\sin\theta\sin\phi)$ where $r \ge 0$, $\theta \in [0,\pi]$ and $\phi \in [0,2\pi)$. The surface becomes
$$(x^2+y^2+z^2)^2 = x \iff r^4 = r\cos\theta$$
Since $r \ge 0$, the relevant range of $\theta$ is $[0,\pi/2]$ where $\cos\theta \ge 0$.
The volume we seek becomes
$$\int_0^{2\pi} \int_0^{\pi/2} \left(\int_0^{\sqrt[3]{\cos\theta}} r^2 dr\right) \sin\theta d\theta d\phi
=\int_0^{2\pi}\int_0^{\pi/2} \left[ \frac13 r^3 \right]_{r=0}^{\sqrt[3]{\cos\theta}} \sin\theta d\theta d\phi\\
= \frac{2\pi}{3}\int_0^{\pi/2} \cos\theta\sin\theta d\theta
= \frac{\pi}{3}
$$
A: For every x, you have a constant $y^2+z^2$. That means for every x we can make a cross section which looks like a circle on a plane parallel to the yz-plane, centered at that x value on the x axis with radius r such that $r^2=y^2+z^2$. If we take the square root in your expression, we get
$$r^2=\sqrt x - x^2$$
There’s no need for a plus or minus because everything is squared and therefore positive.
Using the classic volume of a solid of revolution technique, we get
$$\pi \int_0^1 r^2 dx = \int_0^1 \sqrt x - x^2 dx$$
(We know bound is $x=1$ because for $x>1$, the left hand side is necessarily larger than the right hand side, because the left consists of $x^4+$non negative terms, while the right hand side is just $x$)
A: Try $\int_0^1 \pi r(x)^2\,dx$ where $r(x)$ is obtained from @Aaron's hint.
