Could anyone help with this problem?

Evaluate $\iint_S \textbf{F}$ $\cdot$ $\textbf{n}$ $d \alpha$ using Stoke's Theorem or the Divergence Theorem.

c) S is the truncated cone $y=2\sqrt{x^2+z^2}$, $2 \le y \le 4$, $ \textbf{n}$ is the outward-pointing normal, and $\textbf{F}(x,y,z)=(x,-2y,z)$

From the comments I am guessing the answer depends on whether we include the end caps or not. I am not sure if we have to include them. In class we used the divergence theorem to solve this problem and we got 0, but the back of the book has $28\pi$.

*It's not homework.

  • $\begingroup$ Do you mean $z = 2\sqrt{x^2 + y^2}$, rather than $y$? $\endgroup$ – Jesse Madnick Apr 27 '11 at 5:14
  • $\begingroup$ Are you including the "caps" of the truncated cone? Because that makes a big difference... $\endgroup$ – Jesse Madnick Apr 27 '11 at 5:16
  • $\begingroup$ @Jesse: I think so. We actually solved this problem in class, we got 0, but the back to the book has something else. I am guessing the book did not include the end caps but we did include them in class. $\endgroup$ – user2467 Apr 27 '11 at 5:47

Only one of the theorems (Stokes or Divergence) can apply. Which one applies depends on your situation.

  • Stokes' Theorem applies if your surface has a boundary curve. This would mean that your truncated cone (frustum) does not include the end caps.

  • The Divergence Theorem applies if your surface is closed. This would mean that your frustum includes the end caps.

The way your problem is stated, it would seem as if the end caps are not included, meaning that the Divergence Theorem does not apply. However, while Stokes' Theorem does apply, it's difficult to see how it could be used practically here. My advice would then be to compute the integral directly, probably via a sort of rotated cylindrical coordinate system.

If the end caps are included, then you can use the Divergence Theorem and get an answer of $0$. This is because if $\mathbf{F}(x,y,z) = (x, -2y, z)$, then $\text{div }\mathbf{F} = 1 - 2 + 1 = 0$, so $\iint \mathbf{F}\cdot \mathbf{n}\, dS = \iiint \text{div }\mathbf{F}\,dV = 0$.

As an interesting aside, note that if both of the theorems applied to a surface $S$, then we would have $$\int_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \text{curl }\mathbf{F}\cdot \mathbf{n}\,dS = \iiint_E \text{div}(\text{curl }\mathbf{F})\,dV = 0,$$ and the integral of any vector field over the boundary curve $C$ would be zero. This is a fancy way of saying that "the boundary of a boundary is zero." In other words, if a volume has a boundary surface, then that surface cannot have a boundary curve.

  • $\begingroup$ Actually, in the case where the caps are not included, you can apply Stokes' Theorem so long as you can find a vector field $\mathbf{G}$ such that $\text{curl }\mathbf{G} = \mathbf{F}$. In general, this is not exactly straightforward. However, if you're lucky -- or good at guessing, like I sometimes can be -- you can stumble upon $\mathbf{G}(x,y,z) = (3yz, 2xz, xy)$ and use that, but this is certainly not a reliable strategy in general! $\endgroup$ – Jesse Madnick Apr 27 '11 at 6:30
  • $\begingroup$ Oh! yes, we did talk about that in class. Thank you Jesse. $\endgroup$ – user2467 Apr 27 '11 at 6:35
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    $\begingroup$ Can we also problem by completing the solid, adding the two ends caps, applying the divergence theorem and then solving for the surface integral that we want? Since div F = 0, we would have that the sum of the surface integrals is 0. $\endgroup$ – user2467 Apr 27 '11 at 23:21
  • $\begingroup$ Yes, that would certainly work. Now why didn't I think of that... $\endgroup$ – Jesse Madnick Apr 28 '11 at 3:05

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