# Divergence Theorem - Curve Integrals

1. Mar 19, 2006

This question is throwing me for a loop.

Q: If $u = x^2$ in the square $S = \{ -1<x,y<1\}$, verify the divergence theorem when $\vec w = \Nabla u$:

$$\int\int_S div\,grad\,u\,dx\,dy = \int_C \hat n \cdot grad\,u\,ds$$

If a different $u$ satisfies Laplace's equation in $S$, what is the net flow through C?

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So the first thing I should do is show that each side of the expression match:

$$\vec w = \Nabla u = (2x,0)$$

(1) $$\int\int_S div\,grad\,u\,dx\,dy = \int_{-1}^{1} \int_{-1}^{1} \Nabla \cdot \vec w \, dx \, dy = 2 \int_{-1}^{1} \int_{-1}^{1} \,dx\,dy = 8$$

(2) $$\int_C \hat n \cdot \nabla u \,ds = \sum\limits_{i = 1}^4 \int_C \hat n_i \cdot \vec w(\vec r_i) ds$$

$$\vec w = (2x,0)$$
$$ds = \sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2} dt = \frac{d}{dt}(2(\vec r_{xi}(t)))dt$$

A parametric line is of the form:
$$\vec r(t) = \vec r_0 - t\vec r_0 + t\vec r_1$$

Since we have $\vec w \cdot \hat n_i$ only the x-component of the path is used. And to trace the square we have the parametric paths:

$$r_1(t) = (2t-1,---)$$
$$r_2(t) = (1,---)$$
$$r_3(t) = (-2t+1,---)$$
$$r_4(t) = (-1,---)$$

The normal vectors for each path would be:
$$\hat n_1 = (0,1)$$
$$\hat n_2 = (1,0)$$
$$\hat n_3 = (0,1)$$
$$\hat n_4 = (-1,0)$$

Evaluating:
$$\int_0^1 (1,0)\cdot (2(1),0)2(1)dt + \int_0^1 (-1,0) \cdot (2(-1),0) 2(-1)dt = 4 - 4$$

Now I'm doing something wrong. But to be honest, I feel so lost on this problem it's not funny.

Some of my concerns are as follows:
* I'm not doing this question correctly AT ALL.
* $\vec w$ is conservative. So the fundemental theorem of vector calculus should hold. So going from the start point to the end point which is the same point will yield 0? Or is this when you are only traveling in a vector field.

* This part of the question:
If a different $u$ satisfies Laplace's equation in $S$, what is the net flow through C?

What is C? Is C the path traced around the square?

* Ahhh! :(
If anyone could shed some light on this problem for me, that'd be awesome. Thanks.

Last edited: Mar 19, 2006
2. Mar 19, 2006

Wait a second. $ds$ is supposed to be giving me an arc length. How can I get a negative length returned from it. Ok, so I'm guessing the second part of the curve integral is wrong, so it DOES return 8 like it should.

Also... I grabbed my calc book out and looked at the fundemental theorem again. It is not for dealing with $\int_C f(x,y) \, ds$ integrals. It is for vector fields. Since the curve integral (in the form f(x,y)ds is really returning a sum of all the heights f(x,y) above the x-y plane (assuming a positve function). So again 0 even on a closed path doesn't make any sense.

Ok, I think I cleared that part up anyway.... I hope so :)
Vector calc confuses the hell out of me sometimes, weird notation I guess (just not very comfortable with it yet).

I just need to tackle the part about laplaces equation now. (I hope). So, any help on that would be awesome.

By the way... I still feel rather confused with this question. Which I'm sure you can tell.

3. Mar 20, 2006