Use a triple integral to compute the volume of a solid bounded by yhe surface $$(x^2+y^2+z^2)^3=3xyz$$ I use spherical coordinates \begin{cases} x=r \cos \varphi \cos \theta \\ y=r \sin \varphi \cos \theta \\ z=r\sin \theta \end{cases}
but what I get is $(x^2+y^2+z^2)^3=3xyz \Rightarrow$
$r^6=3r^3\cos \varphi \cos \theta \sin \varphi \cos \theta\sin \theta \Rightarrow$
$r^3=3 \cos \varphi \sin \varphi \cos^2 \theta \sin \theta$
and it doesn't look helpful to me. Can anyone explain?