Evaluating Integral with Green's Theorem: x^2+y^2=9

[MacLaddy]

Homework Statement

Use Green's Theorem to evaluate \int_c(x^2ydx+xy^2dy), where c is the positively oriented circle, x^2+y^2=9

Homework Equations

\int\int_R (\frac{\delta g}{\delta x}-\frac{\delta f}{\delta y})dA

The Attempt at a Solution

I have found \frac{\delta g}{\delta x}-\frac{\delta f}{\delta y} to be y^2-x^2

My hangup is moving forward. My integral will look like this,

\int\int_R [y^2-x^2]dA

however, since the region is a circle I am integrating over should I convert this to polar? If I do, will my values in the integral be rcos^2\theta - rsin^2\theta, or since it's basically just a line integral, will it be 3cos^2\theta - 3sin^2\theta?

This setup is confusing me. Any help is appreciated. Am I integrating just the perimeter of the circle, or the entire thing?

Thanks,
Mac

 

[SammyS]

Some typos / errors corrected below:

Homework Statement

Use Green's Theorem to evaluate \int_c(\red{x}^2ydx+xy^2dy), where c is the positively oriented circle, x^2+y^2=9

Homework Equations

\int\int_R (\frac{\delta g}{\delta f}-\frac{\delta f}{\delta y})dA

Should be

\displaystyle \int\int_R (\frac{\delta g}{\delta x}-\frac{\delta f}{\delta y})dA

The Attempt at a Solution

Following line corrected:

I have found \frac{\delta g}{\delta x}-\frac{\delta f}{\delta y} to be y^2-x^2
My hangup is moving forward. My integral will look like this,

\int\int_R [y^2-x^2]dA

however, since the region is a circle I am integrating over should I convert this to polar? If I do, will my values in the integral be rcos^2\theta - rsin^2\theta, or since it's basically just a line integral, will it be 3cos^2\theta - 3sin^2\theta?

The r's should be squared.

r^2\cos^2(\theta) - r^2\sin^2(\theta)

This setup is confusing me. Any help is appreciated. Am I integrating just the perimeter of the circle, or the entire thing?

Thanks,
Mac

What integral do you get ?

 

[MacLaddy]

Some typos / errors corrected below:
Should be

\displaystyle \int\int_R (\frac{\delta g}{\delta f}-\frac{\delta f}{\delta y})dA

Following line corrected:

Sorry, this is confusing me. I fixed this \int\int_R (\frac{\delta g}{\delta f}-\frac{\delta f}{\delta y})dA to be \int\int_R (\frac{\delta g}{\delta x}-\frac{\delta f}{\delta y})dA this, but it may have been while you were typing. Let me know if there are still errors that I am missing.

The r's should be squared.

r^2\cos^2(\theta) - r^2\sin^2(\theta)What integral do you get ?

Well, if I convert to polar, and from what you are saying, it should be
\displaystyle \int_0^{2\pi}\int_0^3[r^2 sin^2\theta - r^2 cos^2\theta]rdrd\theta
Do I still use the additional r in the Jacobian when I am doing it this way? Or am I completely off base?

Thanks for the help.
Mac

 

[SammyS]

Sorry, this is confusing me. I fixed this \int\int_R (\frac{\delta g}{\delta f}-\frac{\delta f}{\delta y})dA to be \int\int_R (\frac{\delta g}{\delta x}-\frac{\delta f}{\delta y})dA this, but it may have been while you were typing. Let me know if there are still errors that I am missing.

Yes, you must have corrected it while I was typing.

Well, if I convert to polar, and from what you are saying, it should be
\displaystyle \int_0^{2\pi}\int_0^3[r^2 sin^2\theta - r^2 cos^2\theta]rdrd\theta
Do I still use the additional r in the Jacobian when I am doing it this way? Or am I completely off base?

Thanks for the help.
Mac

Of course you use the Jacobian. dxdy → rdrdθ .

 

[MacLaddy]

Of course you use the Jacobian. dxdy → rdrdθ .

Groovy, thank you very much for the help. My mind was definitely glitching on that one.

I evaluated the integral and it equaled 0, so I suppose there is no net rotation.

Mac