Divergence theorem SUPER complex, maybe

[samee]

Homework Statement

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

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.

 

[klondike]

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

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

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.

 

[samee]

In principle, yes. The gradient of a function.

It should be (1/r)(∂/∂r)(rv_r)+(1/r)(∂/∂θ)v_θ+(∂/∂z)v_z

 

[klondike]

Gradient and divergence are different.
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

In principle, yes. The gradient of a function.

It should be (1/r)(∂/∂r)(rv_r)+(1/r)(∂/∂θ)v_θ+(∂/∂z)v_z

 

[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!)

 

[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??

 

[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.

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??

 

[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!

 

[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.

 

[samee]

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

 

[LCKurtz]

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

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.