# Finding the outward flux through a sphere

Problem:

Find the flux of of the field $F$ across the portion of the sphere $x^2 + y^2 + z^2 = a^2$ in the first octant in the direction away from the origin, when $F = zx\hat{i} + zy\hat{j} + z^2\hat{k}$.

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This is just a direct application of a formula, so if you tell me where you are stuck, I'll gladly help you. – Martin Argerami Feb 26 '12 at 0:36
I'm having a hard time getting n hat – Sandy James Feb 26 '12 at 0:56

The way you calculate the flux of $F$ across the surface $S$ is by using a parametrization $r(s,t)$ of $S$ and then $$\int\!\!\!\!\int_S F\cdot n\, dS = \int\!\!\!\!\int_D F(r(s,t))\cdot (r_s\times r_t)\, dsdt,$$ where the double integral on the right is calculated on the domain $D$ of the parametrization $r$.

In this case, since $S$ is a sphere, you can use spherical coordinates and get the parametrization $$r(\theta, \phi)=(a\cos\theta\sin\phi, a\sin\theta\sin\phi, a\cos\phi),\ \ 0\leq\theta\leq\frac\pi2,\ \ 0\leq\phi\leq\frac\pi2.$$ The "first octant" is chosen by the region where we let $\theta$ and $\phi$ vary (if you think carefully about it you'll see that $\pi/2$ is the right choice above).

Now the partial derivatives: $$r_\theta=(-a\sin\theta\sin\phi,a\cos\theta\sin\phi, 0),\ \ \ r_\phi=(a\cos\theta\cos\phi, a\sin\theta\cos\phi, -a\sin\phi).$$ The normal vector: \begin{eqnarray} r_\theta\times r_\phi&=&\left|\begin{matrix}i& j& k\\ -a\sin\theta\sin\phi&a\cos\theta\sin\phi& 0\\ a\cos\theta\cos\phi& a\sin\theta\cos\phi& -a\sin\phi \end{matrix}\right| \\ \ \\ &=&(-a^2\cos\theta\sin^2\phi, -a^2\sin\theta\sin^2\phi, -a^2\sin\phi\cos\phi). \end{eqnarray} Since we want the direction away from the origin, we need to reverse the signs in the normal vector. Now \begin{eqnarray} F(r(\theta,\phi))\cdot(r_\theta\times r_\phi)&=& (a^2\cos\theta\sin\phi\cos\phi,a^2\sin\theta\sin\phi\cos\phi,a^2\cos^2\phi) \\ & &\cdot(a^2\cos\theta\sin^2\phi, a^2\sin\theta\sin^2\phi, a^2\sin\phi\cos\phi) \\ &=& a^4\cos^2\theta\sin^3\phi\cos\phi+a^4\sin^2\theta\sin^3\phi\cos\phi+a^4\sin\phi\cos^3\phi\\ &=& a^4\sin\phi\cos\phi(\cos^2\theta\sin^2\phi+\sin^2\theta\sin^2\phi+\cos^2\phi)\\ &=&a^4\sin\phi\cos\phi. \end{eqnarray} Finally, $$\int\!\!\!\!\int_S F\cdot n\, dS = \int_0^{\pi/2}\!\!\int_0^{\pi/2}a^4\sin\phi\cos\phi\,d\theta d\phi=\frac\pi2\,a^4\left.\frac{\sin^2\phi}2\right|_0^{\pi/2}=\frac{\pi a^4}4$$

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Thank you so much for all of your help, you really saved me! – Sandy James Feb 26 '12 at 15:38

This is $\int_R F \cdot n \,dS$ where $R$ denotes the boundary of portion of the sphere $x^2 + y^2 + z^2 = a^2$ where $x,y,z \geq 0$, because $F \cdot n$ is zero on the flat sides of $R$ and thus the integral over those portions is zero.

By the divergence theorem, the integral is $\int_O div\, F \,dx\,dy\,dz$, where $O$ is the portion of the sphere where $x,y,z \geq 0$. This is $$\int_O 4z \,dx\,dy\,dz$$ Converting to spherical coordinates this is $$\int_0^{\pi \over 2} \int_0^{\pi \over 2}\int_0^a 4\rho^3 \cos(\phi)\sin(\phi)\,d\rho\,d\theta\,d\phi$$ $$= {\pi \over 2}\int_0^a 4\rho^3\,d\rho\int_0^{\pi \over 2}\cos(\phi)\sin(\phi)\,d\phi$$ $$= {\pi a^4 \over 2}\bigg({1 \over 2}\sin^2(\phi)\big|_{\phi = 0}^{\phi = {\pi \over 2}}\bigg)$$ $$= {\pi a^4 \over 4}$$

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I think this is wrong. To apply the divergence theorem you need a closed volume. Which means that what you are really calculating is the flux not only over the part of the sphere, but also on the three sides $x=0$, $y=0$, $z=0$. In this case you just got lucky that those three additional faces contribute nothing because of the particular form of the field $F$. – Martin Argerami Feb 26 '12 at 4:47
I didn't get lucky, I noticed this and then decided to use the divergence theorem. See my first paragraph. By the way, your answer is off by a factor of 2. – Zarrax Feb 26 '12 at 6:57
I missed that sentence, sorry. But it is your answer that is off by a factor of two. You missed the sine from the Jacobian (it is $\rho^2\sin\phi$, and you just put $\rho^2$), and your $\phi$ integrand should have been $\cos\phi\sin\phi$. – Martin Argerami Feb 26 '12 at 14:08
alright, it's been corrected, thanks for pointing that out. – Zarrax Feb 26 '12 at 15:11