0
$\begingroup$

Please help!!!

Demonstrate the identity

$\int_D \nabla \times \mathbf{F} dV = \oint_{\partial D} \mathbf{\hat{n}} \times \mathbf{F} dS$,

by evaluating both sides of the equality for

$\mathbf{F} = xy\mathbf{\hat{k}}$,

where $D$ is the unit cube defined by

$0 \le x \le 1 $ , $0 \le y \le 1$ , $0 \le z \le 1$.

As usual, $\mathbf{\hat{n}}$ is the outward-pointing unit normal to the surface, $\partial D$.

$\endgroup$
4
  • $\begingroup$ What do you need help with specifically? $\endgroup$ – David H May 20 '14 at 23:47
  • $\begingroup$ how to evaluate each integral... I have calculated the curl operator and found it to be $x\hat{i} - y\hat{j} + 0\hat{k}$, but not sure what to do after that... basically evaluating each integral $\endgroup$ – Ameer May 20 '14 at 23:49
  • 1
    $\begingroup$ OK, they are asking you to evaluate the identity with the 'curl' operator; I assumed 'divergence' b/c of title of question. Will change answer. $\endgroup$ – Christopher K May 20 '14 at 23:52
  • $\begingroup$ Alright thanks a bunch! :D $\endgroup$ – Ameer May 20 '14 at 23:53
1
$\begingroup$

For divergence:

Well, we know that $\nabla \cdot F \equiv 0$. Now, $F$ only "leaks" through the surfaces $z=0$ and $z=1$. Why? Also, by independence of $F$ from the variable $z$ we see that the two surface integrals are equal and opposite. Both sides evaluate to $0$.

For curl:

Left hand side is $\int_{D} (x, -y, 0) = (1/2, -1/2, 0)$. On the other side, do not concern yourself with $z=0$ or $z=1$. For $x=1$, we have $$\int_{x=1} (0, -xy, 0) + \int_{x=0} (0, xy, 0) + \int_{y=1} (xy, 0, 0) + \int_{y=0} (-xy, 0, 0) \\ = (0, -1/2, 0) + (0, 0, 0) + (1/2, 0, 0) + (0, 0, 0) \\ = (1/2, -1/2, 0)$$ as desired.

Note:

What is key here is that you have to integrate the components of a vector for the "curl" part. I must admit this seems somewhat "unnatural" at first. Indeed, it threw me for a loop when I saw it for the first time. I will also clarify that by $x=1$ I mean the surface $[1]\times[0,1]\times[0,1]$ and $\int_{x=1} (0, -xy, 0) = \int_{0}^{1}\; \int_{0}^{1}\; (0, -y, 0)\; dy\; dz$, since $x=1$ over the surface.

$\endgroup$
3
  • $\begingroup$ Thank you!!!!!!!!!!!!!!!!!! $\endgroup$ – Ameer May 21 '14 at 0:02
  • $\begingroup$ @Ameer To select this answer as the correct one, click on the green check mark next to the answer. +1, by the way, Chris! $\endgroup$ – user122283 May 21 '14 at 0:12
  • 1
    $\begingroup$ @SanathDevalapurkar, thank you. Ameer, if you are satisfied with my answer, then I'll gladly accept your stamp of approval. If, on the otherhand, you want to see other posts, feel free to keep it open. $\endgroup$ – Christopher K May 21 '14 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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