Finding the Correct Function and Limits for a Polar Double Integral

[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 where R is the region http://ada.math.uga.edu/webwork2_files/tmp/equations/21/9cc2c610adbc73bdcbe1922b3dea321.png

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

 

[Office_Shredder]

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

 

[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.

 

[glid02]

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

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

 

[Max Eilerson]

He is using degrees rather than radians?

 

[glid02]

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

 

[TMFKAN64]

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

 

[glid02]

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