(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement, all variables and given/known data

http://s1.ipicture.ru/uploads/20120121/YUmpPl4N.jpg

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.

[tex]\hat{n}=x\vec{i}+y\vec{j}+z\vec{k}[/tex][tex]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[/tex][tex]d\sigma=\frac{1}{\sqrt{1-x^2-y^2}}\,.dxdy[/tex]

Converting to polar coordinates:

[tex]\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[/tex]

Let's just say i spent a certain amount of time doing the integration.

[tex]\int^1_0 \frac{r^3 {\cos}^2 \theta}{\sqrt{1-r^2}}\,.dr=\frac{2{\cos}^2 \theta}{3}[/tex]

[tex]\int^1_0 2r^2\sin\theta\,.dr=\frac{2\sin\theta}{3}[/tex]

[tex]\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}[/tex]

[tex]\int^{\frac{\pi}{2}}_0 \frac{2{\cos}^2 \theta}{3}+ \frac{2\sin\theta}{3}= \frac{\pi}{6}+ \frac{2}{3}[/tex]

Somehow i got it wrong, as the correct answer is [itex]\frac{\pi}{6}[/itex]

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# Homework Help: Evaluate outward flux

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