What is the volume of that body? The formula of the surface is 
$$\left({x \over{a}}\right)^{\frac 23} + \left({y \over{b}}\right)^{\frac 23} + \left({z \over{c}}\right)^{\frac 23} = 1$$
where all the factors are positive numbers.
The main problem is that I have no idea how does this surface looks like. So probably that's the reason why all substitutions that I have tried didn't help at all. I thought it was a good idea to use cylindrical coordinates
$$
\begin{cases}
x = r \cdot \cos(\phi) \\
y = r \cdot \sin(\phi) \\
z = z
\end{cases}
$$
with Jacobin $J = r$, but it also didn't work.
Thanks for any help!
 A: The equation 
$$\left({x \over{a}}\right)^{\frac 23} + \left({y \over{b}}\right)^{\frac 23} + \left({z \over{c}}\right)^{\frac 23} = 1$$
does not change by putting $-x $ for $x $, $-y $ for $y $ and $-z $ for $z $, hence the surface is symmetric in all eight octants. Thus volume of entire solid= $8\times $Volume in first octant.   
But volume of solid at any point $(x,y,z) $ is $dxdydz $. Hence, volume in positive octant $=\int\int\int dxdydz  $ where the integral is extended to all positive values of the variables subject to $(\frac {x}{a})^{2/3} +(\frac {y}{b})^{2/3} +(\frac {z}{c})^{2/3} $.   
Putting $(\frac {x}{a})^{2/3} =u, (\frac{y}{b})^{2/3} =v, (\frac {z}{c})^{2/3} =w \Rightarrow x=au^{3/2}, y=bv^{3/2}, z=cw^{3/2} $.    
Hence, the volume $= \int\int\int \frac {27}{8} abc u^{3/2-1}v^{3/2-1}w^{3/2-1} dudvdw$ where $u+v+w\leq 1$. Thus, volume= $\frac {27}{8}abc\frac {[\Gamma(3/2)]^3}{\Gamma (\frac {9}{2}+1)} $ which can be easily simplified to $\frac {\pi abc}{8}\frac{4}{35} $.   
Total volume=$8$ times volume in first octant =$\frac {4}{35}(\pi abc) $. Hope it helps.
