Find $\iint_S\vec{F}\cdot\vec{n} dS$

Question

Find $\iint_S\vec{F}\cdot\vec{n} dS$ where $\vec{F}=x\vec{i}+y\vec{j}+z\vec{k}$ and S is the boundary surface of the region bounded by the cone $z=\sqrt{x^2+y^2}$ and the upper half sphere $x^2+y^2+z^2=8$. ($\vec{n}$ is outward pointing normal).

My attempt:

Consider $z=\sqrt{x^2+y^2}$ and $x^2+y^2+z^2=8$ I got $x^2+y^2=4$ which means radius 2.

We know that $\iint_S\vec{F}\cdot\vec{n} dS=\iint_S\vec{F}\cdot dS=\iint_D\vec{F}\cdot (r_u\times r_v)dA=\iint_D(-P\frac{\partial g}{\partial x}-Q\frac{\partial g}{\partial y}+R)dA$

So $\iint_S\vec{F}\cdot\vec{n} dS$ becomes $\iint_D-\frac{x^2}{\sqrt{x^2+y^2}}-\frac{y^2}{\sqrt{x^2+y^2}}+\sqrt{x^2+y^2}dA$

Using polar coordinate, it becomes $\int_0^{2\pi}\int_0^2rdrd\theta$ and I finally got $4\pi$

I don't have the answer but I think I should use stoke's theorem or divergence theorem on this question. Please help.

Answer 1

Yes, in this case you may find the result directly, but the final result is different $$\begin{align*}\iint_S\vec{F}\cdot\vec{n} dS&=\iint_{\text{Cone}}\vec{F}\cdot\vec{n} dS+\iint_{\text{Cap}}\vec{F}\cdot\vec{n} dS\\ &=0+\sqrt{8}\text{Area(Cap)}\\ &=\sqrt{8}(2\pi\sqrt{8}(\sqrt{8}-2))\\ &=32\pi(\sqrt{2}-1) \end{align*}$$ where the first integral is $0$ because $\vec{F}$ is orthogonal to $\vec{n}$ along the cone and the second one is $\sqrt{8}\text{Area(Cap)}$ because $\vec{F}$ is parallel to $\vec{n}$ along the spherical cap with constant $|\vec{F}|=\sqrt{8}$.

P.S. Note that $$\begin{align*}\text{Area(Cap)}&=\int_{x^2+y^2\leq 2^2}\sqrt{1+z_x^2+z_y^2}\,dxdy\\ &= 2\pi\int_{0}^2\frac{\sqrt{8}}{\sqrt{8-r^2}}\,rdr\\ &=2\pi\sqrt{8}\left[-\sqrt{8-r^2}\right]_0^2\\ &=2\pi\sqrt{8}(\sqrt{8}-2)\end{align*}$$ where $z(x,y):=\sqrt{8-x^2-y^2}$.