Question 66 for GR9768

Forum for the GRE subject test in mathematics.
moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Question 66 for GR9768

Postby moo5003 » Tue Oct 14, 2008 4:15 pm

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?

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm

Postby CoCoA » Tue Oct 14, 2008 4:28 pm

A circle with center at zero is a very nice object in polar. Try converting the double integral of (2x+2)dA to polar.

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby moo5003 » Tue Oct 14, 2008 4:38 pm

Thanks, completley forgot about polar. Made it much easier ^_^.

ngoc
Posts: 6
Joined: Mon Nov 03, 2008 5:54 am

Postby ngoc » Tue Nov 04, 2008 12:41 am

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?

ngoc
Posts: 6
Joined: Mon Nov 03, 2008 5:54 am

Postby ngoc » Tue Nov 04, 2008 12:55 am

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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