Polar double integral

1. Nov 1, 2006

glid02

I'm having trouble finding the function and/or the limits to this problem:

Using polar coordinates, evaluate the integral http://ada.math.uga.edu/webwork2_files/tmp/equations/01/19aeef09224e0fca11ef9d6e45fb311.png [Broken] where R is the region http://ada.math.uga.edu/webwork2_files/tmp/equations/21/9cc2c610adbc73bdcbe1922b3dea321.png [Broken]

I've tried the function as sin(r^2) with the limits as 3 to 6 and 0 to 2pi, but that does not give the right answer.

That would give -1/2*cos(r^2) which gives .0893 dtheta, which is then
.0893*2pi.

I'm pretty sure I'm doing the integrals correctly, I think I just have the function wrong. If anyone can help me out I'd appreciate it.

Thanks

Last edited by a moderator: May 2, 2017
2. Nov 1, 2006

Office_Shredder

Staff Emeritus
You realize you need to change it from dA to $$drd\theta$$ right?

Last edited: Nov 1, 2006
3. Nov 1, 2006

TMFKAN64

$$r dr d\theta$$. But he did that, and -1/2*cos(r^2) looks right to me. Your limits look good too. I don't see where you are getting .0893 though.

Last edited: Nov 1, 2006
4. Nov 1, 2006

glid02

I evaluated 1/2*cos(r^2) from 6 to 3

-1/2(cos(36)-cos(9))

Last edited: Nov 1, 2006
5. Nov 1, 2006

Max Eilerson

He is using degrees rather than radians?

6. Nov 1, 2006

glid02

It's in radians, all the other homework has been and I just tried the answer with 360 instead of 2pi.

7. Nov 1, 2006

TMFKAN64

Max meant that your calculation used degrees, when you should have used radians... and I think he's right.

8. Nov 1, 2006

glid02

Ah, you are right. That worked, thanks a lot.

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