Find volume with triple integral

Question

I want to find the volume of $ (x^2 + y^2)^3 + z^6 =a^3xyz $.

I changed the basis like that: $ x = r \sin ^ \frac{1}{3} B \cos A; y = r \sin ^ \frac{1}{3} B \sin A; z = r \cos ^ \frac{1}{3} B $ (similar to spherical coordinates), so the equalation changed to $ r^3 = a^3 \sin ^ \frac{2}{3} B \cos ^ \frac{1}{3} B \cos A \sin A $. But when I took integral with $A=0..2\pi; B=0..\frac {\pi} {2}; r=0.. (a^3 \sin ^ \frac{2}{3} B \cos ^ \frac{1}{3} B \cos A \sin A)^\frac {1} {3}$ and Jacobian $= \frac {-r^2 \cos {2b} }{3 \cos ^ \frac{2}{3} B \sin ^ \frac{1}{3} B}$ I got $0$ as the answer. Would be nice if you showed me the mistake or another method, thanks.

Answer 1

For $0\le A\le\frac{\pi}{2}$, and $\pi\le A\le \frac{3\pi}{2}$, $r\ge 0$, For the other half of the $A$ domain, $r\le 0$. Therefore the $A$ integration has to be split up into four parts so that each part integrates to a positive result.

Caveat: I am still not sure what the figure looks like.

Answer 2

Why would you set $x = r \sin ^ \frac{1}{3} B \cos A$? What does that even mean, and what does it have to do with spherical coordinates?

I would expect something like $x = r \sin B \cos A.$