# Stokes's Theorem

1. Feb 16, 2008

### EugP

1. The problem statement, all variables and given/known data
For the vector field $$\bold{E} = \bold{ \hat x} (xy) - \bold{ \hat y} (x^2 + 2y^2)$$, calculate the following:

$$\oint \bold{E} \cdot d\bold{l}$$ around the triangular contour shown.

I don't have a scanner at the moment so I will explain the drawing. The picture is a right triangle. They show an x and y axis. From (0, 0), there is a line going to (1, 0), then from there it goes up to (1, 1), then a diagonal back to (0, 0).

2. Relevant equations

3. The attempt at a solution

I know how to approach it but I seem to be stuck. This is what I have so far:

I gave each coordinate a name: a (0, 0), b (1, 0), and c (1, 1).

$$\oint \bold{E} \cdot d\bold{l} = \oint_a^b \bold{E}_{ab} \cdot d\bold{l} + \oint_b^c \bold{E}_{bc} \cdot d\bold{l} + \oint_c^a \bold{E}_{ca} \cdot d\bold{l}$$

At this point I'm stuck. Could someone please point me in the right direction?

2. Feb 16, 2008

### jhicks

Your confusion probably lies in that you don't know what $$\vec{dl}$$ is. For example, for the integral from a to b, $$\vec{dl_{1}}=\hat{x}dl$$, whereas b to c it becomes $$\vect{dl_{2}}=\hat{y}dl$$. Just work out the dot products, which are especially easy for a to b and b to c since either x or y is constant.

Last edited: Feb 16, 2008
3. Feb 16, 2008

### EugP

Thanks for the quick reply. I have 2 questions:

1) Shouldn't the integral from a to b, be $$\oint_a^b \bold{E}_{ab} \cdot \hat x d\bold{l}$$ since ab is horizontal?

2) I'm not sure what to put for $$\bold{E}_{ab}$$. Should it be the component that acts only in the direction of ab, in other words:

$$\oint_a^b \bold{ \hat x} (xy) \cdot \hat x d\bold{l}$$

Am I in the right direction?

4. Feb 16, 2008

### jhicks

Yes whoops you are right about 1). I will change it. And 2) yes you are right. Don't forget that y is a constant.

5. Feb 16, 2008

### EugP

Awsome. So now that I set it up, I have $$\oint_a^b xy d\bold{l}$$, but I don't know l, or am I missing something obvious?

6. Feb 16, 2008

### jhicks

Remember that a is (0,0) and b is (1,0). $$dl$$ is simply dx in that particular case. The only tricky part is going to be c to a, where you're going to have to express all components of dl in terms of one variable to integrate over.

Last edited: Feb 16, 2008
7. Feb 16, 2008

### kdv

If the path is along the positive x axis, you must use $$d \vec{l} = dx \vec{i}$$

and so on.

In general, $$d\vec{l} = dx \vec{i} + dy \vec{j} + dz \vec{k}$$

If you have a parametrized curve, you replace all the variables in terms of the parameter and the integral is only over one variable,

8. Feb 16, 2008

### EugP

Alright I got it. Thanks for all the help guys.

9. Feb 16, 2008

### HallsofIvy

Staff Emeritus
Since you titled this "Stokes Theorem", are you also going to integrate -2x-2y over the triangular region and show that they are the same?

10. Feb 18, 2008

### EugP

Actually yes, and I'm stuck on that too.

Also, I thought I figured out how to finish the first part, but it's not working out.

Here's what I'm getting:

$$\oint \bold{E} \cdot d\bold{l} = \oint_0^1 xy dx + \oint_0^1 -x^2-2y^2 dy + \oint_c^a \bold{E}_{ca} \cdot d\bold{l}$$
but I don't know what to do with the last term. And when I integrate the first two terms, I get
$$\frac{y}{2}-x^2-\frac{2}{3}$$

which doesn't make sense, because the answer for the whole is -1. What should I do with the last term? Shouldn't it be:

$$\oint_0^1 (xy - x^2 +2y^2) dxdy$$

11. Feb 19, 2008

### Defennder

I got -1 for both methods. Did you get -1 for the line integral?

12. Apr 20, 2009

### cartonn30gel

this is a perfect question to use Green"s theorem. And the answer is -1. Stokes' theorem is just a generalized version of Green's th. and I don't see how that would be useful in this case. Also there is no need to do 3 integrations. Just convert the whole thing to a double integral using Green's theorem, integrate over the triangle and get -1