1. Area between two circles

Find the area of the region between the circles x^2 + y^2 = 4 and x^2 + y^2 = 4x

My attempt at the solution:
I found the points of intersection are at (1,-sqrt(3)) and (1, sqrt(3))
Then I solved for x in each equation:
x=sqrt(4-y^2) , x=sqrt(y^2+4) + 4

This is where I get stuck. I don't know where to go from here. Can someone please give me a hint or suggestion? Also, is what I Have so far correct?

3. Re: Area between two circles

Hello, nubshat!

Find the area of the region between the circles: $\displaystyle x^2 + y^2 \,=\, 4\,\text{ and }\,x^2 + y^2 \,=\,4x$

The graph looks like this . . .

Code:
                |
* * *     * * *
*     |   * *         *
*       | *::|//*         *
*        |*:::|///*         *
|::::|/A//
*         *::::|////*         *
--*---------*----+----*---------*--
*         *::::|1:::*2        *
|::::|::::
*        |*:::|:::*         *
*       | *::|::*         *
*     |   * *         *
* * *     * * *
|
Note that region $\displaystyle A$ is one-fourth of the desired area

and $\displaystyle A$ can be found with: .$\displaystyle \int^2_1\sqrt{4-x^2}\,dx$

5. Re: Area between two circles

Post #3 is correct. The integral expressions should be swapped (and 2 added) in post #2: WolframAlpha.

It is easy to find the area without calculus. Using notations from post #2, we need to find twice the area of the circular segment PLRQ. Then $\displaystyle \tan\angle POL=\sqrt{3}$, so the angle is 60 degrees. Therefore, the central angle POR of the segment is 120 degrees. The area of the corresponding circular sector is $\displaystyle (120/360)\pi r^2$, and the area of the triangle POR is $\displaystyle (1/2)PO\cdot OR\sin\angle POR =\sqrt{3}$. From here, the area of PLRQ is $\displaystyle (4/3)\pi-\sqrt{3}$, and the final answer is twice that.