Area inside r = 2 cos(θ) but outside r = 1

[filter54321]

Homework Statement

Use double integrals to find the area inside the circle r = 2 cos(θ) and outside the circle r = 1.

Homework Equations

I figured this was too easy to require an graphic. If you can't picture the circles, imagine them in rectangular from:
r = 2 cos(θ) ==> y²+(x-2)²=1
r = 1 ==> y²+x²=1

The Attempt at a Solution

Both circles have a radius of 1 and you need to look at all 2\pi of the objects to see the full area of overlap. So this is what I tried:

\int\stackrel{2\pi}{0}\int\stackrel{0}{1} (2cos(θ)-r) drdθ

The book says the answer is but I can't get it:

\frac{\pi}{3}+\frac{\sqrt{3}}{2}

 

[Just a nobody]

The integral to find area in polar coordinates is:
\iint_A r \, dr \, d\theta

Adjust the limits of integration to match the equations given. The actual contents of the integral (r \, dr \, d\theta) will remain the same.

 

[LCKurtz]

You need to find the polar coordinates of the two curves intersections and use appropriate limits. Not everything goes from 0 to 2\pi or 0 to 1.

Generally to find an area using polar coordinate double integrals you need something like this:

A = \int_{\theta_1}^{\theta_2} \int_{r_{inner}}^{r_{outer}}1\ rdrd\theta

and you need to determine the correct limits from your formulas and picture.

 

[filter54321]

Thanks, but that doesn't really get me any closer to an answer. Is the function I'm integrating correct? What are the ranges?

 

[LCKurtz]

Thanks, but that doesn't really get me any closer to an answer. Is the function I'm integrating correct? What are the ranges?

I can't make much sense out of your integrals. To get area with a double integral, you integrate the function 1. You need to look at the graphs. Find the \theta's where they intersect by setting the r values equal.