I am working on the following exercise:
Calculate the volume of the body bounded by the following surface:
$$(x^2+y^2+z^2)^2 = xyz$$
I would solve this with a triple integral, but I do not see how I can derive the boundaries from $(x^2+y^2+z^2)^2 = xyz$. I guess one way to do this would be to use spherical coordinates, i.e we substitute
$$x = r \cdot \sin(\theta) \cos(\varphi), y = r \cdot \sin(\theta) \sin(\varphi), z = \cos(\theta) $$
, but I can't take it further from there. Could you help me?
Mea maxima culpa. The previous version of this post calculated the area incorrectly. I've given a hopefully correct calculation of the volume below.
We have (as you can check) $$ r^4 = r^3 \sin^2 \theta \cos \theta \sin \phi \cos\phi $$ So $$ r = \sin^2 \theta\cos\theta \sin\phi \cos\phi \equiv f(\theta , \phi) $$ at the boundary of the volume. Following the advice of Empy2, we compute the volume in the first octant: \begin{align} V_1 &= \int_{0}^{\pi/2} \int_{0}^{\pi/2} \int_0^{f(\theta, \phi)} r^2 \sin \theta \ dr \, d\theta \, d\phi \\&= \int_{0}^{\pi/2} \int_{0}^{\pi/2}\frac{(f(\theta, \phi))^3}{3} \sin \theta \ d\theta \, d\phi \\&= \frac{1}{3}\int_0^{\pi/2} \sin^7\theta \cos^3 \theta \ d\theta \int_0^{\pi/2} \sin^3 \phi \cos^3 \phi \ d\phi \\&= \frac{1}{12}\int_0^1 u^3(1 - u) \ du \int_0^{1} v(1 - v) \ dv \\&= \frac{1}{12} \left(\frac{1}{4} - \frac{1}{5} \right)\left(\frac{1}{2} - \frac{1}{3} \right) \\&= \frac{1}{12} \cdot \frac{1}{20} \cdot \frac{1}{6} = \frac{1}{1440} \end{align} We notice that the defining equation of the surface has no solutions with an odd number of the coordinates $(x,y,z)$ negative (the LHS is always positive, so the RHS cannot be negative). It's also interesting to note that if any one of the coordinates of a solution $(x,y,z)$ to the equation is $0$, then the rest must also be $0$ to have equality. Finally, observe that if any two of the coordinates of a solution $(x,y,z)$ are multiplied by $-1$, then the resulting point is also a solution. This shows that the surface consists of $4$ identical lobes intersecting only at the origin, with each lobe contained entirely in an octant with an even number of positive coordinates. Therefore, the total volume is $$ V = 4 V_1 = \frac{1}{360} $$