Calculating Triple Integral

Question

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).

Answer 1

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 $$8\int _0^1\int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}27 a b c \rho ^8 \cos ^2(\theta ) \cos ^2(\varphi ) \sin ^2(\theta ) \sin ^5(\varphi ) \rho ^2d\varphi d\theta d\rho = 8\frac{9}{770} \pi a b c.$$

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

Answer 2

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....