Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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 $F = zx\hat{i} + zy\hat{j} + z^2\hat{k}$.

share|cite|improve this question
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
up vote 4 down vote accepted

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, $$ \int\!\!\!\!\int_S F\cdot n\, dS = \int_0^{\pi/2}\!\!\int_0^{\pi/2}a^4\sin\phi\cos\phi\,d\theta d\phi=\frac\pi2\,a^4\left.\frac{\sin^2\phi}2\right|_0^{\pi/2}=\frac{\pi a^4}4 $$

share|cite|improve this answer
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 $$\int_0^{\pi \over 2} \int_0^{\pi \over 2}\int_0^a 4\rho^3 \cos(\phi)\sin(\phi)\,d\rho\,d\theta\,d\phi$$ $$= {\pi \over 2}\int_0^a 4\rho^3\,d\rho\int_0^{\pi \over 2}\cos(\phi)\sin(\phi)\,d\phi$$ $$= {\pi a^4 \over 2}\bigg({1 \over 2}\sin^2(\phi)\big|_{\phi = 0}^{\phi = {\pi \over 2}}\bigg)$$ $$= {\pi a^4 \over 4}$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.