<s>[INST] How find this integral $I(a,b)=\iiint_{x^2+y^2+z^2\le 1}(ax+by)^2\,dx\,dy\,dz$ Find this integral
$$I(a,b)=\iiint_{x^2+y^2+z^2\le 1}(ax+by)^2\,dx\,dy\,dz$$
since
$$(ax+by)^2=a^2x^2+b^2y^2+2abxy$$
so
$$I=I_{1}+I_{2}+I_{3}$$
where
$$I_{1}=\iiint_{x^2+y^2+z^2\le 1}ax^2\,dV=a\iiint_{x^2+y^2+z^2\le 1}x^2\,dV$$
since
$$\iiint_{x^2+y^2+z^2\le 1}x^2\,dV=\iiint_{x^2+y^2+z^2\le 1}y^2\,dV$$
maybe have other methods? Thank you
 [/INST] 
With (almost) no computations:

The domain $x^2+y^2+z^2\leqslant1$ is invariant by the rotations of the plane $(x,y)$ hence $$I(a,b)=(a^2+b^2)\,J,$$ where
$$
J=\int\!\!\!\iint_{x^2+y^2+z^2\leqslant1}x^2\mathrm dx\mathrm dy\mathrm dz.
$$
By symmetry with respect to the $(x,y,z)$ axes, $3J=K(1)$, where
$$
K(t)=\int\!\!\!\iint_{x^2+y^2+z^2\leqslant t^2}(x^2+y^2+z^2)\mathrm dx\mathrm dy\mathrm dz.
$$
By differentiation, using the fact that the surface of the sphere of radius $t$ is $4\pi\,t^2$,
$$
K'(t)=4\pi\,t^2\cdot t^2=4\pi\,t^4.
$$
Since $K(0)=0$, for every nonnegative $t$,
$$
K(t)=\frac45\pi\,t^5,
$$
and, finally,
$$
I(a,b)=\frac13\,\frac45\pi\,(a^2+b^2).
$$
</s>