Use the divergence theorem to evaluate the surface intergral $$\iint_S\vec{A}.\vec{n}dS$$ where we know that $$\nabla .\vec{A}=4r\cos^2(\theta)$$ where $S$ is the spherical surface of unit radius centered on the origin.

For my attempt i get $$\iint_S\vec{A}.\vec{n}dS=\int_{r=0}^{r=1}\int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\frac{\pi}{2}}4r\cos^2(\theta)dV$$ By the divergence theorem where i think that $$dV=r^2\sin(\theta)dr d\theta d\phi $$ For my answer i get $\frac{2\pi}{3}$ but my lecturer has just put his answer only online which he got was $\frac{4\pi}{3}$ i dont understand where i'm going wrong if i am going wrong, i dont undesrtand where he is getting the factor of $2$ from. Any help would be appreciated, Thanks.


Given your answer, I am pretty sure you are confusing two things: First, which angle is represented by which variable. Given your jacobian and the correct answer of $4\pi/3$, I believe you mean $\theta$ to be the angle sweeping from $k$ to $-k$. Second, spherical coordinates require this $\theta$ sweep from $0$ to $\pi$, not $\pi/2$.

Thus, your integral should be $$ \int_0^1\int_0^{\pi}\int_0^{2\pi} 4r^3\cos^2(\theta)\sin(\theta)\mathrm d\phi \mathrm d\theta \mathrm dr= 2\pi \int_0^\pi\cos^2(\theta)\sin(\theta) \mathrm d\theta=\frac{4\pi}{3} $$

edit: from context (the answer given by the lecturer) I am going to assume that $\theta$ is the angle sweeping along the z axis.

  • $\begingroup$ Thank you! how can one calculate dV or is it just something to remember? $\endgroup$ – user395952 May 7 '17 at 19:08
  • $\begingroup$ Not a problem. You can absolutely calculate it. It is called the jacobian and is given by the determinant of the derivative of a change of coordinates, here your change of coordinates is the change to spherical coordinates $\endgroup$ – qbert May 7 '17 at 19:11
  • $\begingroup$ is it possible you could show me this, i'm not used to the jacobian but would like to learn how to do it would be easier than remembering all the different variations of dV, and also is this the same for dS when doing surface intergrals? $\endgroup$ – user395952 May 7 '17 at 19:18
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    $\begingroup$ @Gibberish these notes: tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx are always excellent $\endgroup$ – qbert May 7 '17 at 19:23

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