I want to find a volume of a body, bounded by:

$$(x^2+y^2+z^2)^2 = az(x^2+y^2)$$ for some $a > 0$.

As I understood I am supposed to say that for аor a fixed value of z, it looks like a circle, which means that we just need to take the integral of the resulting function.

But I'm having a little trouble getting the radius $(x ^ 2 + y ^ 2)$ of a circle at fixed z and reducing that to an integral.

Maybe there are simpler ways to get the volume of a given shape?

## Best Answer

In spherical coordinates, your equality becomes$$\rho^4=a\rho\cos(\varphi)\rho^2\sin^2(\varphi)=a\rho^3\cos(\varphi)\sin^2(\varphi).$$So, your volume is\begin{align}\int_0^{2\pi}\int_0^{\pi/2}\int_0^{a\cos(\varphi)\sin^2(\varphi)}\rho^2\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta&=2\pi\int_0^{\pi/2}\frac13a^3\cos^3(\varphi)\sin^7(\varphi)\,\mathrm d\varphi\\&=\frac{a^3\pi}{60}.\end{align}