# Is this integral set up correctly?

1. Nov 24, 2006

### G01

Find the area of the region within both circles;

r=cos(theta) and
r=sin(theta)

using a double integral.

$$2\int^{\pi/2}_{\pi/4} \int^{\cos\theta}_0 r dr d\theta$$

I multiplied by 2 because the area I have is only half of the total area to be found. Is this correct or am i doing something stupid?

2. Nov 24, 2006

### NateTG

I'm not sure that it's a particularly good example for using multiple integrals, but you're effectively using Green's Theorem on something that can be readily handled by:
$$A = \frac{1}{2} \int r^2 d\theta$$
or, for those who prefer not to deal with polar integrals:
$$A=2 \int_0^{\frac{1}{2}} \left(\sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}-x \right)dx$$

I'm fairily sure that your expression will give a numerically correct result.

http://mathworld.wolfram.com/GreensTheorem.html

Last edited: Nov 24, 2006
3. Nov 24, 2006

### G01

Thanks Nate, I agree that its probably easier to do this with
$$1/2 \int r^2 d\theta$$ but our assignment was to do it using a double integral. Thanks for the help. I'll keep working on it and see if i get the right answer.