# Double integrals using polar coordinates

1. Nov 11, 2009

### popo902

1. The problem statement, all variables and given/known data

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles
x^2 + y^2 = 4
and
x^2 - 2x + y^2 = 0

2. Relevant equations

3. The attempt at a solution

for my integral i got
0<= theta <=pi/2 for the theta bounds since it lies in the 1st quad
and i got
2costheta<= r <= 2 for the bounds of r because 2costheta is the polar version of
x^2 - 2x + y^2 = 0 and 2 is 'r' in the equation x^2 + y^2 = 4

and the equation that i integrated is (2- 2costheta)r
you add the extra r when turning the equation polar and the equation r=2 is over
r = costheta when i graphed it
and welll...
i got pi - 3 but it's wrong...

am i even setting up the equation right or am i just having problems with the math?

2. Nov 12, 2009

### Staff: Mentor

This is a double integral, so what you want is this:
$$\int_{\theta = 0}^{\pi/2}~\int_{r = ?}^{?} r~dr~d\theta$$

IOW, the limits of integration describe the region, and all you need for the integrand is 1dA. You seem to have a handle on how to describe the region, so use what you have found for the limits for r.

I get $\pi/2 - 1/2$ for the area.

3. Nov 15, 2009

### popo902

r u serious??
why is it do i integrate 1dA?
i saw that in the book for finding the area of one of the clovers of a cosine function
but it didn't explain why i just integrate 1dA

4. Nov 15, 2009

### Staff: Mentor

Yes, I'm serious. For a region R, this double integral gives the area:
$$\int \int_R 1 dA$$

In rectangular coordinates, the iterated integral will look something like this:
$$\int_{y = a}^b \int_{x = f(y)}^{g(y)} 1 dx~ dy$$

and dA = dx dy or dA = dy dx. If you integrate in the reverse order, the limits of integration will be different.

In polar coordinates, dA = r dr d$\theta$. That's where the r came from.

5. Nov 17, 2009

### popo902

oh right. I'm looking for the area.
double integral of 1da gives the area and triple integral of 1 gives the volume of a 3d object.
thank you

6. Nov 17, 2009

### Staff: Mentor

double integral of 1dA gives the area and triple integral of 1dV gives the volume of a 3d object.