= Solution
On the radius-$a$ sphere, the outward unit <normal vector> is $(x,y,z)/a$, so $\mathbf F\cdot\mathbf n=(x^2+2y^2+3z^2)/a$. For a direct <surface integral>, put $x=a\sin\theta\cos\varphi$, $y=a\sin\theta\sin\varphi$, $z=a\cos\theta$ and $dS=a^2\sin\theta\,d\theta\,d\varphi$. The integral is
$$
a^3\int_0^{2\pi}\!\int_0^\pi[\sin^2\theta(\cos^2\varphi+2\sin^2\varphi)+3\cos^2\theta]\sin\theta\,d\theta\,d\varphi.
$$
The azimuthal integrals are $\pi,\pi,2\pi$, while $\int_0^\pi\sin^3\theta\,d\theta=4/3$ and $\int_0^\pi\cos^2\theta\sin\theta\,d\theta=2/3$. Thus
$$
\boxed{\int_S\mathbf F\cdot d\mathbf S=a^3\left[3\pi\frac43+6\pi\frac23\right]=8\pi a^3.}
$$
The <flux of a linear field through a sphere> also follows from the equality of the three spherical second moments, without using the <divergence theorem>.
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