# Area between two circles

• Nov 16th 2013, 06:36 PM
nubshat
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
Attachment 29750
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?
• Nov 16th 2013, 07:21 PM
ibdutt
Re: Area between two circles
• Nov 16th 2013, 07:41 PM
Soroban
Re: Area between two circles
Hello, nubshat!

Quote:

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: .$\displaystyle \int^2_1\sqrt{4-x^2}\,dx$
• Nov 16th 2013, 08:48 PM
nubshat
Re: Area between two circles
• Nov 17th 2013, 08:40 AM
emakarov
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.