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

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.

• Yes, you should use the Divergence Theorem and turn this into a triple integral over the region they described. That is almost always easier than doing surface integrals. Commented Dec 1, 2019 at 19:28

## 1 Answer

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 $$z(x,y):=\sqrt{8-x^2-y^2}$$.

• Why is it $\frac{\sqrt8}{\sqrt{8-r^2}}$? Commented Dec 1, 2019 at 15:53
• $\sqrt{1+z_x^2+z_y^2}$ should give us $\sqrt2$ Commented Dec 1, 2019 at 15:57
• Oh I just notice what the z is Commented Dec 1, 2019 at 15:59
• @BrianWu Any further doubt? Commented Dec 2, 2019 at 8:04