Use a triple integral to compute the volume of a solid bounded by yhe surface $(x^2+y^2+z^2)^3=3xyz$ 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?
 A: You can graph the parametric surface on a function grapher. I did the correction because there are some mistakes in the answers.
If you had graphed the surface, you will see that the volumen is divided in four 
polygons in diferents quadrants. I will find the volumen when $x,y,z\geq0$.
$$(x^2+y^2+z^2)^3=3xyz$$
Use de Jacobian for change to spherical coordinates
\begin{cases}
x=r \cos \varphi \cos \theta \\
y=r \sin \varphi \sin \theta \\
z=r\cos \varphi
\end{cases}
Then, $(x^2+y^2+z^2)^3=3xyz \Rightarrow$
$r^3=3 \sin^2 \varphi \cos \theta \sin \theta \cos\varphi$
We will obtain the following when calculating the Jacobian.
$$r^2\sin\varphi$$
We define the volume to be integrated.
$$D:={(x,y,z)/0\leq\theta\leq\pi/2;0\leq\varphi\leq\pi/2;0\leq r\leq\sqrt[3]{3\sin^2 \varphi \cos \theta \sin \theta \cos\varphi}}$$
Because I will take one of four polygons in the firts cuadrant.
$$ 4\int_{0}^{\pi/2}d\theta\int_{0}^{\pi/2}d\varphi\int_{0}^{\sqrt[3]{3\sin^2 \varphi \cos \theta \sin \theta \cos\varphi}}r^2\sin\varphi dr =4\int_{0}^{\pi/2}d\theta\int_{0}^{\pi/2}\sin^3 \varphi \cos \theta \sin \theta \cos\varphi=\int_{0}^{\pi/2}cos \theta \sin \theta=\color{red}{\frac{1}{2}}$$
