Line Integrals

1. Mar 15, 2004

wubie

Hello,

I must be having some sort of brain malfunction or something. First here is my question:

Evaluate the line integral with respect to arc length

The integral sub C of x*e^y ds where C is the arc of the unit circle from (1,0) to (-1,0) traversed counterclockwise.

Now if the circle is traversed counterclockwise then

-1 <= x <= 1 and 0 <= y <= 1.

I can parameterize the equation by noticing that

x = cos theta and y = sin theta.

Therefore the integral becomes

The integral of cos theta * e^sin theta ds

where ds is

( (-sin theta)^2 + (cos theta)^2 )^1/2 = 1

So the integral becomes

The integral of cos theta * e^sin theta dtheta.

Since the unit circle is traversed counterclockwise the angle goes from 0 to pi.

If I let sin theta = u then du = cos theta and the limits become zero and zero.

So if I integrate the integral of e^u du then I get

e^u with the limits of 0 and 0 which ends up to be zero.

This cannot be right can it? I am confused. I think I did everything correctly. Where am I going wrong?

Perhaps I am right?

Any help is appreciated. Thankyou.

2. Mar 15, 2004

Looks right to me, but then I've always sucked at path integrals, so. Maybe Matt will grace you with his presence.

It actually kinda seems right, too. It's an odd function with respect to the y-axis, and since we're going from -1 to 1, we have a symmetric domain. That suggests an answer of 0.

3. Mar 15, 2004

HallsofIvy

Staff Emeritus
Yes, that's exactly right. "xey" is an odd function with respect to x and since the path of integration is symmetric with respect to the y-axis the integral is 0.

If it makes you feel better, you might try looking at exactly the same problem from (0,-1) to (0,1) instead. You get exactly the same integral except that theta now runs from -pi/2 to pi/2. Since you are making the substitution u= sin(theta), you integration will be from
-1 to 1 giving a non-zero result. That is, of course, correct since the function is not symmetric with respect to the x-axis.