# Divergence theorem SUPER complex, maybe

1. Jul 10, 2012

### samee

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

use divergence theorem to evaluate ∫s∫F dot n dA if

F=[sinh yz, 0, y4] , S: r=[u,cosv,sinv], -4≤u≤4 , 0≤v≤pi

3. The attempt at a solution

Instructor surprised us with this one, I have no idea how to attempt. I know that ∫vdiv v dV=∫sn dot v dA, which is the divergence theorem, but I have no idea how to apply it.

2. Jul 10, 2012

### klondike

Do you know how to find $$\nabla \bullet F$$

3. Jul 10, 2012

### samee

In principle, yes. The gradient of a function.

It should be (1/r)(∂/∂r)(rvr)+(1/r)(∂/∂θ)vθ+(∂/∂z)vz

4. Jul 10, 2012

### klondike

One takes gradient of scalar field. The result is vector field.
One takes divergence of vector field. The result is scalar field.

You are given a vector function F in cartesian coordinates. The following link should help. http://en.wikipedia.org/wiki/Divergence#Application_in_Cartesian_coordinates

5. Jul 10, 2012

### HallsofIvy

Since your function as given in terms of x, y, and z, wouldn't it be better to use
$$\nabla\cdot f\vec{i}+ g\vec{j}+ h\vec{k}= \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$$
be better?

Now, what is the divergence of this vector function, $sinh(yz)\vec{i}+ y^4\vec{k}$?

(Quite frankly, this problem is remarkably trivial!)

6. Jul 10, 2012

### samee

OK!

So (∂f/∂x)=(∂/∂x)sinh(yz)=0
and (∂h/∂z)=(∂/∂z)y4=0

so the divergence of F is zero! and anything dot zero is zero, right? so the answer is zero??

7. Jul 10, 2012

### klondike

Looks good.

Whenever one sees a crazy closed surface integral or contour line integral, it would be wise to check if the vector function is divergence free or curl free respectively.

8. Jul 10, 2012

### samee

Thank you so much! I took an incomplete due to family issues in this class and I'm trying to make it up this summer, but it's been such a mess trying to remember the 1st half of the class and do homework for the 2nd half with almost a year between. Thank you for your help!!!

9. Jul 10, 2012

### LCKurtz

Hold on a minute there everyone. The divergence theorem applies to a surface enclosing a volume, which this isn't. You can add three more obvious surfaces to your given surface $S$ to enclose volume. Call them $S_1,\,S_2,\, S_3$. Now if your surface is oriented so your enclosed volume surface is directed outward (orientation wasn't given), you can apply the divergence theorem. So what you have so far is$$\iint_{S_1\cup S_2 \cup S_3 \cup S}\vec F \cdot d\vec S=\iiint_V \nabla\cdot \vec F\ dV = 0$$
That just tells you the integral over $S$ is minus the sum of the surface integrals over the other 3 surfaces. You aren't finished until you check them.

10. Jul 10, 2012

### samee

Okay, so I check them out by computing the dot product of F over each surface?

11. Jul 10, 2012

### LCKurtz

Yes. What you hope for in this type of problem is that the surface integral over the other three surfaces is easier to calculate than the original surface integral. I think you will find that is the case in this problem.