# 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 where R is the region

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

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.