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 Post subject: Question 66 for GR9768
PostPosted: Tue Oct 14, 2008 4:15 pm 
Joined: Mon Oct 06, 2008 7:33 pm

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Joined: Mon Oct 06, 2008 7:33 pm
Posts: 36
66. The line integral [x^2 dy -2y dx] over the circle x^2+y^2=9.

Now I tried using Greene's theorem on this but got bogged down when doing the double integral.

M = -2y
N = x^2

Integral of 2x + 2 dA.

First integrate with respect to y with limits +/- Root(9-x^2)

Then you get the integral from -3 to 3 of:

4(x+1)Root(9-x^2) <--- I stopped here.

Instead of using Greene’s theorem I started over parameterizing the equation in terms of t and then solving it directly using a few trig substitutions along the way (answer was 18pi). I was wondering what you guys think the best way to tackle this problem is?


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 Post subject:
PostPosted: Tue Oct 14, 2008 4:28 pm 
Joined: Wed Sep 03, 2008 5:39 pm

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Joined: Wed Sep 03, 2008 5:39 pm
Posts: 42
A circle with center at zero is a very nice object in polar. Try converting the double integral of (2x+2)dA to polar.


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 Post subject:
PostPosted: Tue Oct 14, 2008 4:38 pm 
Joined: Mon Oct 06, 2008 7:33 pm

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Joined: Mon Oct 06, 2008 7:33 pm
Posts: 36
Thanks, completley forgot about polar. Made it much easier ^_^.


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 Post subject:
PostPosted: Tue Nov 04, 2008 12:41 am 
Joined: Mon Nov 03, 2008 5:54 am

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Joined: Mon Nov 03, 2008 5:54 am
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Even with polar I still get $36\pi$ (twice the answer). I'm not sure why??

Also, I doubled check my answer by reasoning that:
\int_C (-y)dx = area of the circle = 9\pi.
Hence 2\int_C (-y)dx = 18\pi
and the left over is:
\int x^2 dy, (which is not a conservative vector field), and in fact can be calculated to be 18\pi. So the ansewr is 36\pi again. Please tell me where I went wrong?


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 Post subject:
PostPosted: Tue Nov 04, 2008 12:55 am 
Joined: Mon Nov 03, 2008 5:54 am

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Joined: Mon Nov 03, 2008 5:54 am
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Ah ok -- never mind. I figured out that I made a mistake in both methods (polar coord with Green's theorem, AND integration directly). So 18\pi it is. :)


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