# Using stokes' or divergence theorem to solve integral

1. Oct 11, 2013

### leroyjenkens

1. The problem statement, all variables and given/known data

Use either Stokes' theorem or the divergence theorem to evaluate this integral in the easiest possible way.

∫∫V $\cdot$ndσ over the closed surface of the tin can bounded by x2+y2=9, z = 0, z = 5, if V = 2xyi - y2j + (z + xy)k

The bolded letters are vectors.

2. Relevant equations

∫∫( xV) $\cdot$ ndσ = ∫V $\cdot$dr Stokes' theorem

∫∫∫ $\cdot$Vdτ = ∫∫V $\cdot$ndσ Divergence theorem

3. The attempt at a solution

So what I did was use the divergence theorem to turn this into a volume integral. I did the ∇xV to get 2y-2y+1. But here's where I'm stuck. I can input the formula for the volume of a cylinder, right? Is that dτ? I have to integrate with respect to τ. So if 2y-2y+1 = 1, then I'm left with τ after I integrate, and is τ just ∏r2h?

Thanks

2. Oct 11, 2013

### vela

Staff Emeritus
You mean you calculated $\nabla\cdot \vec{V}$, not the curl. $d\tau$ is the volume element. For cartesian coordinates, it's equal to $d\tau = dx\,dy\,dz$. The fact that you're integrating over a cylindrical volume is reflected in the limits. If I were you, however, I'd switch to cylindrical coordinates, where $d\tau = r\,dr\,d\theta\,dz$.

EDIT: Never mind the last sentence. I misread your expression for the divergence and thought you would need to actually evaluate a non-trivial integral. Your work looks fine. If the divergence weren't simply a constant, you'd probably be better off switching to cylindrical coordinates.

3. Oct 11, 2013

### Zondrina

That all sounds reasonable. You can switch to cylindrical co-ordinates to make integrating easier I believe.

4. Oct 11, 2013

### Dick

But the divergence isn't a constant. It's 2x-2y+1. After you integrate over the cylinder then the 2x and 2y drop out. But you should say why.

5. Oct 11, 2013

### vela

Staff Emeritus
That's the same mistake I made originally, but the first term is really 2y, not 2x.

6. Oct 11, 2013

### Dick

Ooo. Yes, it is.