# Evaluate outward flux

1. Jan 20, 2012

### DryRun

The problem statement, all variables and given/known data

The attempt at a solution
First of all, taking a look at the wording in the problem itself. I am not sure if the unit normal vector is supposed to be outward or inward. There is the "outward flux" but the "inside the sphere" part is contradicting. Anyway, i went with "outward", although i'm not 100% sure.
$$\hat{n}=x\vec{i}+y\vec{j}+z\vec{k}$$$$Flux=\iint \vec{F}.\hat{n}\,.d \sigma=\iint x^2+2yz\,.d\sigma=\iint x^2+2y\sqrt{1-x^2-y^2}\,.d\sigma$$$$d\sigma=\frac{1}{\sqrt{1-x^2-y^2}}\,.dxdy$$
Converting to polar coordinates:
$$\iint x^2+2y\sqrt{1-x^2-y^2}\,.d\sigma=\iint \frac{x^2}{\sqrt{1-r^2}}+2y\,.rdrd\theta=\iint \frac{r^3 {\cos}^2 \theta}{\sqrt{1-r^2}}+2r^2\sin\theta\,.drd\theta$$
Let's just say i spent a certain amount of time doing the integration.
$$\int^1_0 \frac{r^3 {\cos}^2 \theta}{\sqrt{1-r^2}}\,.dr=\frac{2{\cos}^2 \theta}{3}$$
$$\int^1_0 2r^2\sin\theta\,.dr=\frac{2\sin\theta}{3}$$
$$\int^1_0 \frac{r^3 {\cos}^2 \theta}{\sqrt{1-r^2}}+2r^2\sin\theta\,.dr=\frac{2{\cos}^2 \theta}{3}+\frac{2\sin\theta}{3}$$
$$\int^{\frac{\pi}{2}}_0 \frac{2{\cos}^2 \theta}{3}+ \frac{2\sin\theta}{3}= \frac{\pi}{6}+ \frac{2}{3}$$
Somehow i got it wrong, as the correct answer is $\frac{\pi}{6}$

Last edited: Jan 20, 2012
2. Jan 20, 2012

### Dick

You can't use the divergence theorem? If I'm reading that right, you spent a lot of time finding the flux through the spherical part of the octant. The boundary of the octant also has planar parts. Integrating through all of the parts isn't the easiest way to do this.

Last edited: Jan 20, 2012
3. Jan 21, 2012

### DryRun

After some more thinking, i realize that
$$dV=\frac{1}{8}\times \frac{4\pi r^2}{3}$$
where r=1, so $$dV=\frac{\pi}{6}$$
div F=1, so the answer is $$\frac{\pi}{6}$$
I can't believe it was so easy...