# Why can't I get Stoke's Theorem to Work ?

1. Sep 25, 2007

### Orson1981

Why can't I get Stoke's Theorem to Work!!!?

1. The problem statement, all variables and given/known data
This is very similar to the problem I put up last night actually part c of the same problem. I really have put an amazing amount of time into this problem (week and a half/ 5 hrs a day) and it countinues to stump me. At this point I'm spinning mental wheels and could use some help.

we have a 3-4-5 right triangle lying on the x-y plane with the length three side on the x-axis and the length 4 side on the y-axis. we define a field
$$\vec{u}=x^2\hat{x} + xy\hat{y}$$

and I am trying to evaluate the integral:

$$\oint{\vec{u}d\vec{l}}$$

2. Relevant equations

I need to solve both:

$$\oint{\vec{u}d\vec{l}}$$

and

$$\oint{(\nabla{}x\vec{u}) . \hat{n} dA}$$
please excuse my clumsy notation, this should be del cross the vector u

3. The attempt at a solution

The surface integral after doing del cross u I end up with

$$\frac {1}{2} \oint{ y \hat{k} dxdy}$$
evaluated from 0 - 3 on the x-axis and 0 - 4 on the y-axis
Doing this I get 12.
the 1/2 I added in to this equation is largely a guess on my part, and could be correct or not, just seemed to make some sense for a triagle who's area is 1/2 base*height

Looking at the line integral

I placed curve 1 along the y axis from 4 - 0 then paramaterized x = 0 and dx = 0
so for curve 1 is zero

Curve 2, I paramaterized y = 0, dy = 0 that left

$$\int{x^2 dx}$$ evaluated from 0 - 3 gives me 9

Curve 3 along the hypotinous (I can't spell ... sorry) paramaterized y = 4 - 4/3 * x dy = -3/4 * dx
plugging these new values into y and dy and integrating from 3 - 0 gives me -1

So for the whole curve I have 8

So somewhere I missed something because last I checked 8 =! 12, or if you ignore my guess of 1/2 in the surface integral 8 =! 24
so I'm only off by a factor of 3 somewhere, though it could be coincidence that's a pretty nice number to be off by, makes me think I'm just missing a term somewhere.
If I had to guess I would say somehow $$\hat{n}$$ is somehow equal to 1/3, or perhaps 1/6 to include my guess about 1/2 in the surface integral

Thank you for reading this long rambling thing, and thanks for any help

-Mo

2. Sep 26, 2007

### Avodyne

I believe your line integrals are all correct [though you have a typo: dy = -(4/3)dx, not -(3/4)dx].

But your surface integral is wrong. You can't just multiply by a factor of one half. You need to integrate y over the area of the triangle. This integral takes the form

$$\int_a^b dx \int_c^d dy \; y$$

where you need to figure out what the limits a, b, c, and d should be so that you are covering the triangle. Key point: c and d may depend on x.

When I do this integral with the correct limits, I get 8, matching your result from the line integrals.

3. Sep 26, 2007

### Orson1981

Thank you!!!!, armed with your hint I went back and 'carefully' looked at one of my undergrad math books I'm using as a reference, and of course there it was staring me in the face the whole time, but until you came along I was just to dense to see it.

and thank you for not giving me the answer, and letting me work it out myself.

I kinda feel bad, because I know these are such easy questions I'm throwing out and I should know this stuff, but I'm stoked now, I can do integrals - Awesome.

4. Sep 26, 2007

### Avodyne

Glad to help. The easy stuff is sometimes the hardest to see when you're in the middle of a problem.