Calculating Stokes's Theorem on a Triangular Contour

[EugP]

Homework Statement

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).

Homework Equations

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?

 

[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.

 

[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?

 

[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.

 

[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?

 

[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.

 

[kdv]

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?

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,

 

[EugP]

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

 

[HallsofIvy]

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?

 

[EugP]

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?

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

 

[Defennder]

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

 

[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