Green's Theorem for Evaluating Line Integrals on Curves

[joemama69]

Homework Statement

Evaluate the \int xyi - xj dot dr over the curve y = 1 - x² from 1,0 to 0,1

Homework Equations

The Attempt at a Solution

I used greens theorem

\int -1 - x dy dx

dy is from 0 to 1-x²

dx is from 0 to 1

\int-x³ + x² - x - 1 dx = -1.4166666666666666666

is this correct

 

[Dick]

How can you use Green's theorem? The curve isn't closed. Just do the line integral.

 

[phsopher]

Well, I suppose he could use Green's theorem by completing the loop through the axes and subtracting their contribution which seems to be zero.

Are you sure the sign of the x³ term is correct though?

 

[joemama69]

"just do the line integral"

\intgradf dot dr = \int xyi - xj dot dr

Is this correct to say...I can say that xyi -xj is grad f as long as i can find the f, and then i would plug the two points into f to get the answer

 

[phsopher]

No, this only applies to conservative vector fields, that is fields whose curl is 0. The vector field you have (xyi -xj) is not conservative and therefore the function you describe doesn't exist. Try finding it and you will get contradicting terms from different derivatives. You can't just plug the end points in because the integral depends on which path you take between those points.

But you were going in the right direction in your first post, just remember that Green's Theorem applies to closed curves.

 

[joemama69]

ok

the -x³ should be positive and thus the answer is -11/12

But how do i make it a closed curve and what do i subtract

 

[phsopher]

You have a curve that takes you from (1,0) to (0,1), which means that you need to add another curve that takes you from (0,1) to (1,0), preferably one that would be easier to integrate over than the original. Then you'd have a closed loop and you could use Green's Theorem. But the answer you get won't correspond to your original integral so you'll have to subtract the portion that you added in the beginning. What would be the easiest way to close the curve in your example?

Note that by calculating the area the way you did you already chose the curve that completes the loop.

Of course you can just calculate the original integral by itself as Dick suggested. For that you would need to parametrize x and y using a single variable so that your integral becomes one-dimensional.

 

[joemama69]

I think i should do it as a line integral over a paramaterized curve

y = 1-x²

r(t) = ti + (-t² +1)j
r'(t) = i + -2tj

xyi - xj

F(r(t) = (-t³ + t)i - tj
F(r(t) dot r'(t) = -t³ + 2t² + t

-\int-t³ + 2t² + t = -11/12

is this correct

 

[Dick]

I think i should do it as a line integral over a paramaterized curve

y = 1-x²

r(t) = ti + (-t² +1)j
r'(t) = i + -2tj

xyi - xj

F(r(t) = (-t³ + t)i - tj
F(r(t) dot r'(t) = -t³ + 2t² + t

-\int-t³ + 2t² + t = -11/12

is this correct

Yes, that's the easy way to do it. But why did you change the sign?? Oh, I see, x goes from 1 to 0, not 0 to 1. That's why isn't it?

 

[phsopher]

t goes from 1 to 0

 

[joemama69]

is that just a cusidence that i got the same answer when i used greens theorem incorectly

 

[phsopher]

Yes, it so happens that the contribution of the curve that you should have used to complete the loop is zero. Basically what you calculated using GT was an integral of xyi - xj first along y = 1 - x² from (1,0) to (0,1), then from (0,1) to (0,0) along the y-axis, then from (0,0) to (1,0) along the x-axis thus coming to the point you started from. However, the latter two integrals are zero so the only contribution comes from the first one.