How to solve this seemingly simple triple integral? $$\iiint_D x^2+y^2+z^2\,dxdydz$$ $D$ is bound by $x=0, y=0,z=0$ and $x+y+z=a$, calculated by rote, I got $\frac{a^5}{20}$, is there any simpler way to do this? I tried using spherical coordinates, but the problem doesn't seem to be simplified.
 A: Here's at least one simplification. Since the integrand $f(x,y,z)=x^2+y^2+z^2$ and the region of integration $D$ are symmetric with respect to interchanging any of pair $x,y,z$, the integral simplifies to any of the following three choices:
$$\begin{align}
\iiint_{D}(x^2+y^2+z^2)\,\mathrm{d}x\mathrm{d}y\mathrm{d}z
&=3\iiint_{D}x^2\,\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
&=3\iiint_{D}y^2\,\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
&=3\iiint_{D}z^2\,\mathrm{d}x\mathrm{d}y\mathrm{d}z.
\end{align}$$
The easiest of the three choices to integrate depends on the chosen order of integration. If you choose to integrate first over $z$, then over $y$, and last over $x$, then you probably want to choose the first option, because this leaves the first two integrations fairly trivial:
$$\begin{align}
\iiint_{D}f(x,y,z)\,\mathrm{d}V
&=3\int_{0}^{a}\mathrm{d}x\int_{0}^{a-x}\mathrm{d}y\int_{0}^{a-x-y}\mathrm{d}z\,\color{blue}{x^2}\\
&=3\int_{0}^{a}\mathrm{d}x\,\color{blue}{x^2}\int_{0}^{a-x}\mathrm{d}y\int_{0}^{a-x-y}\mathrm{d}z.
\end{align}$$
