Calculating volume bounded by area

Question

The problem is calculation of the volume bounded by $(x^2+y^2+z^2)^2=\frac{64}{x^2+y^2}$

I tried using cylindrical coordinates: $x=r*cos(t) ; y=r*sin(t); z=z$

then this gives me:

$(r^2+z^2)^2=\frac{64}{r^2} \Rightarrow (r^2+z^2)r=8 \geq r^3 \Rightarrow r\leq2$

and $z^2=\frac{8}{r}-r^2$

then the integral should be

$\displaystyle \int_0^2 \int_0^{2\pi} \int_{-\sqrt{\frac{8}{r}-r^2}}^{\sqrt{\frac{8}{r}-r^2}} r dzdtdr$

which leads to $4\pi\int_0^2\sqrt{8r-r^4}dr$

I appreciate any hints how to calculate this integral if this is correct so far. Thank you.

Answer 1

Ok, i used spherical coordinates. now the integral is: $\displaystyle \int_0^{2 \pi} \int_{- \pi /2}^{ \pi /2} \int_0^{ a} r^2cos(\theta) drdtd\theta$ where $\displaystyle a=\frac{2}{(cos(\theta))^{\frac{1}{3}}}$and the result is $\displaystyle\frac{16\pi^2}{3}$.

I got the same solution from wolfram alpha for my first try but I'm still wondering how to calculate the integral above.