$$\left( \frac{x}{a} + \frac{y}{b} \right)^2 + \left(\frac{z}{c} \right)^3 \le 1$$
$$x \ge 0, y \ge 0, z \ge 0$$
So I made the following substitution using spherical coordinates
$x = ar\cos^2{\phi}\sin{\theta}$
$y = br\sin^2{\phi}\sin{\theta}$
$z = cr^{\frac{2}{3}}cos^{\frac{2}{3}}{\theta}$
where $\theta \in [0, \pi]$, $\phi \in [0, 2\pi]$
I think I chose the correct substitution because I get that $r^2 \le 1$, but here's the problem. Regarding the boundaries of the angles, if I substitute into $x,y,z \ge 0$ I get
$\cos^2{\phi}\sin{\theta} \ge 0$
$\sin^2{\phi}\sin{\theta} \ge 0$
$\cos^{\frac{2}{3}}{\theta} \ge 0$
From the last equation I can deduce that the boundaries for $\theta$ are $ \theta \in [0, \frac{\pi}{2}]$
Let's try the same for $\phi$ in the first two equations:
Since $\theta$ is in the first quadrant, both cosine and sine are positive. Taking a look at $\phi$, because each function containing $\phi$ is squared and thus is positive for any angle in its domain, does that mean we take the natural boundaries for $\phi$, which are $\phi \in [0, 2\pi]$?