Calculating Triple Integral I have task : find volume of body limited by surface $(\frac{x}{a})^{2/3} + (\frac{y}{b})^{2/3} + (\frac{z}{c})^{2/3}$ = 1. I know that this task is about triple integral. But i have confused by such amazing surface. I also plotted graphic in Mathematica, however it didn't help. Maybe there is a substitution (to сylindrical coordinate system).
 A: Let's write
\begin{align}
  x &= a(\rho\sin(\phi)\sin(\theta))^3 \\
  y &= b(\rho\sin(\phi)\cos(\theta))^3 \\
  z &= c(\rho\cos(\phi))^3,
\end{align}
for then, $(x/a)^{2/3} + (y/b)^{2/3} + (z/c)^{2/3} = \rho^2$. Furthermore, the Jacobian of change of variables is
$$648 \rho ^8 \sin ^2(\theta ) \cos ^2(\theta ) \sin^5(\varphi ) \cos ^2(\varphi ).$$
Thus, the volume can be computed as 
$$8\int _0^1\int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi
   }{2}}27 a b c \rho ^8 \cos ^2(\theta ) \cos
   ^2(\varphi ) \sin ^2(\theta ) \sin ^5(\varphi )
   \rho ^2d\varphi d\theta d\rho = 8\frac{9}{770} \pi  a b c.$$

You can also use the parametrization to visualize the object.  Here's what it looks like for $a=b=3$ and $c=2$.

A: If you were integrating over the volume
$$\left(\frac{x}{a}\right)^{2} + \left(\frac{y}{b}\right)^{2} + \left(\frac{z}{c}\right)^{2} \leq 1$$
you would use spherical polars with $x = ar\sin\theta\cos\phi$, $y = br...$, $z = cr...$.
Now try and modify those so they fit your shape by taking the appropriate powers. Then calculate the Jacobian....
