Calculate $\iiint x^2+y^2+z^2dxdydz$ given $x^2+y^2+z^2<x+y+z$
I tried using substitution $x=rcos\alpha cos\beta+1/2$, $x=rsin\alpha cos\beta+1/2$, $x=rsin\beta+1/2$ It it possible to calculate it using this substitution but it seems rather complicated. I wonder if there is better approach.
By the transformations $(x,y,z)\mapsto(y,z,x)\mapsto(z,x,y)$ leaving the integration domain $D$ unchanged, the given integral equals $$ \iiint_{\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{1}{2}\right)^2\leq \frac{3}{4}}3x^2 \,d\mu =\iiint_{a^2+b^2+c^2\leq\frac{3}{4}}3\left(a+\frac{1}{2}\right)^2\,d\mu$$ or $$ \int_{-\sqrt{3}/2}^{\sqrt{3}/2}3\left(a+\frac{1}{2}\right)^2\cdot \mu\left\{(b,c)\in\mathbb{R}^2: b^2+c^2\leq \frac{3}{4}-a^2\right\}\,da $$ that simplifies to: $$ 3\pi \int_{-\sqrt{3}/2}^{\sqrt{3}/2}\left(a+\frac{1}{2}\right)^2\left(\frac{3}{4}-a^2\right)\,da = \color{red}{\frac{3\sqrt{3}\pi}{5}}.$$
Such integral is also the moment of inertia of a solid sphere rotating around the origin, hence it can be computed through the parallel axis theorem, too.