Calculate volume of area enclosed by the surface $\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}=\sqrt[3]{a^2}$. 
Calculate the volume of the area enclosed by the surface
  \begin{equation}
x^{(2/3)}+y^{(2/3)}+z^{(2/3)} = a^{(2/3)}
\end{equation}
  where $a > 0$ is a constant.

However I'm not sure where to begin.
 A: $$
V: (\frac{x_1}{a})^{2/3}+(\frac{x_2}{a})^{2/3}+(\frac{x_3}{a})^{2/3}\le1$$
$$y_i=(\frac{x_i}{a})^{1/3}$$
$$\frac{D(x_1,x_2,x_3)}{D(y_1,y_2,y_3)}=27a^3y_1^2y_2^2y_3^2$$
$$V=\int_V\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3=\int27a^3y_1^2y_2^2y_3^2\mathrm{d}y_1\mathrm{d}y_2\mathrm{d}y_3$$
$$\left\{
\begin{array}{ccc}
y_1&=&r\cos(\theta_1)\\
y_2&=&r\sin(\theta_1)\cos(\theta_2)\\
y_3&=&r\sin(\theta_1)\sin(\theta_2)
\end{array}
\right .$$
$$\frac{D(y_1,y_2,y_3)}{D(r,\theta_1,\theta_2)}=r^2\sin(\theta_1)$$
So
$$V=\int^{\pi/2}_0\mathrm{d}\theta_1\int^{\pi/2}_0\mathrm{d}\theta_2\int_0^1 27a^3(r\cos(\theta_1))^2(r\sin(\theta_1)\cos(\theta_2))^2(r\sin(\theta_1)\sin(\theta_2))^2r^2\sin(\theta_1)\mathrm{d}r\\
=\int^{\pi/2}_0\mathrm{d}\theta_1\int^{\pi/2}_0\mathrm{d}\theta_2\int_0^1 27a^3r^8\cos^2(\theta_1)\sin^5(\theta_1)\cos^2(\theta_2)\sin^2(\theta_2)\mathrm{d}r$$
$$V=3a^3\frac{\Gamma(3/2)\Gamma(6/2)}{2\Gamma(9/2)}\frac{\Gamma(3/2)\Gamma(3/2)}{2\Gamma(6/2)}=3/4a^3\frac{(\Gamma(3/2))^3}{\Gamma(9/2)}=\frac{a^3}{70}\pi$$
A: I want to remind you the answer above is incorrect, altough almost correct.
The volume calcuted there is 1/8'th of the actual volume, because 'xyz' lets both angles only range from $0$ to $\pi/2$, which means he only accounts for the 1 first octant. 
Because the volume is symmetrical in all three axis, you can just multiply the outcome by 8, or calculate the integral
\begin{align*}
&3a^3\int_0^{\pi}\int_0^{2\pi}\cos^2\theta\sin^2\theta\cos^2\varphi\sin^5\varphi \text{ d}\theta\text{ d}\varphi\\
={}& \dfrac{4a^3}{35}\pi
\end{align*}
(PS See you in college tonight ;))
