Green's Theorem

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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

Last edited:

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SammyS
Staff Emeritus
Homework Helper
Gold Member
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 ?

Last edited:
Gold Member
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

Last edited:
SammyS
Staff Emeritus
Homework Helper
Gold Member
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 .

Gold Member
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