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Green's Theorem

  • Thread starter MacLaddy
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  • #1
MacLaddy
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Homework Statement



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

Homework Equations



[tex]\int\int_R (\frac{\delta g}{\delta x}-\frac{\delta f}{\delta y})dA[/tex]


The Attempt at a Solution



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

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

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

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 [itex]rcos^2\theta - rsin^2\theta[/itex], or since it's basically just a line integral, will it be [itex]3cos^2\theta - 3sin^2\theta[/itex]?

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:

Answers and Replies

  • #2
SammyS
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Some typos / errors corrected below:

Homework Statement



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

Homework Equations


[tex]\int\int_R (\frac{\delta g}{\delta f}-\frac{\delta f}{\delta y})dA[/tex]
Should be

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

The Attempt at a Solution

Following line corrected:
I have found [itex]\frac{\delta g}{\delta x}-\frac{\delta f}{\delta y}[/itex] to be [itex]y^2-x^2[/itex]
My hangup is moving forward. My integral will look like this,

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

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 [itex]rcos^2\theta - rsin^2\theta[/itex], or since it's basically just a line integral, will it be [itex]3cos^2\theta - 3sin^2\theta[/itex]?
The r's should be squared.

[itex]r^2\cos^2(\theta) - r^2\sin^2(\theta)[/itex]
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:
  • #3
MacLaddy
Gold Member
289
8
Some typos / errors corrected below:
Should be

[itex]\displaystyle \int\int_R (\frac{\delta g}{\delta f}-\frac{\delta f}{\delta y})dA[/itex]

Following line corrected:
Sorry, this is confusing me. I fixed this [itex]\int\int_R (\frac{\delta g}{\delta f}-\frac{\delta f}{\delta y})dA[/itex] to be [itex]\int\int_R (\frac{\delta g}{\delta x}-\frac{\delta f}{\delta y})dA[/itex] 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.

[itex]r^2\cos^2(\theta) - r^2\sin^2(\theta)[/itex]What integral do you get ?
Well, if I convert to polar, and from what you are saying, it should be
[tex]\displaystyle \int_0^{2\pi}\int_0^3[r^2 sin^2\theta - r^2 cos^2\theta]rdrd\theta[/tex]
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:
  • #4
SammyS
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Sorry, this is confusing me. I fixed this [itex]\int\int_R (\frac{\delta g}{\delta f}-\frac{\delta f}{\delta y})dA[/itex] to be [itex]\int\int_R (\frac{\delta g}{\delta x}-\frac{\delta f}{\delta y})dA[/itex] 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
[tex]\displaystyle \int_0^{2\pi}\int_0^3[r^2 sin^2\theta - r^2 cos^2\theta]rdrd\theta[/tex]
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 .
 
  • #5
MacLaddy
Gold Member
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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
 

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