Calculate surface integral with force Let $S=\{(x,y,z)\in \mathbb{R} : x^2+y^2+z^2 = 11\}$ and $F(x,y,z)=(2\sqrt{11}x,y^2,z^2)$.
Calculate
$$\int\int_SF(x,y,z)\;dS$$
I've calculated S parameterized, whose expression is:
$$S(x,y) = (x,y,\sqrt{11-x^2-y^2})$$
Hence, $$r_x=\frac{\partial S}{\partial x}=(1,0,\frac{-x}{\sqrt{11-x^2-y^2}})$$ $$r_y=\frac{\partial S}{\partial y}=(0,1,\frac{-y}{\sqrt{11-x^2-y^2}})$$ $$||r_x \times r_y|| = \frac{\sqrt{11}}{\sqrt{11-x^2-y^2}}$$
So $dS = \frac{\sqrt{11}}{\sqrt{11-x^2-y^2}}dxdy$ and then
$$\int\int_SF(x,y,z)\;dS = \int\int_SF(x,y,z)\;\frac{\sqrt{11}}{\sqrt{11-x^2-y^2}}dxdy$$
But then I don't know how to continue with this procedure.
Another idea I have is to establish the following relationship:
$$\int\int_SF\;dS= \int\int\int_V \nabla F\;dV$$
Any hint?
Thanks in advance.
 A: Parameterizing $S$ in spherical coordinates makes this easier, though not by much.
$$\vec r(u,v)=\sqrt{11}(\cos u\sin v,\sin u\sin v,\cos v)$$
with $0\le u\le2\pi$ and $0\le v\le\pi$.
Then take the normal vector to $S$ to be
$$\vec r_v\times\vec r_u=11(\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v)$$
where this order of partial derivatives is taken to make the normal vector point outward, away from the origin.
Then the integral is
$$\begin{align*}
\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S&=\int_0^\pi\int_0^{2\pi}\vec F(\vec r(u,v))\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv\\[1ex]
&=\int_0^\pi\int_0^{2\pi}(22\cos u\sin v,11\sin^2u\sin^2v,11\cos^2v)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv
\end{align*}$$
The resulting integral is a bit complicated, but doable by hand. Of course, resorting to the divergence theorem makes this significantly easier to compute:
$$\begin{align*}
\iint_S\vec F\cdot\mathrm d\vec S&=\iiint_{x^2+y^2+z^2\le11}2\sqrt{11}+2y+2z\,\mathrm dV\\[1ex]
&=\int_0^\pi\int_0^{2\pi}\int_0^{\sqrt{11}}(\sqrt{11}+2\rho\sin\theta\sin\varphi+2\rho\cos\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi
\end{align*}$$
Either way, you would get

 $$\frac{986\pi}3$$

