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?
Hello, nubshat!
Find the area of the region between the circles:
The graph looks like this . . .
Note that region is one-fourth of the desired areaCode:| * * * * * * * | * * * * | *::|//* * * |*:::|///* * |::::|/A// * *::::|////* * --*---------*----+----*---------*-- * *::::|1:::*2 * |::::|:::: * |*:::|:::* * * | *::|::* * * | * * * * * * * * * |
and can be found with: .
Which one of these is correct? I'm getting two different answers...
integral of sqrt(4-x^2) from 0 to 1 + (2* integral of sqrt(4x-x^2) from 1 to 2) - Wolfram|Alpha
4 * integral of sqrt(4-x^2) from 1 to 2 - Wolfram|Alpha
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 , 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 , and the area of the triangle POR is . From here, the area of PLRQ is , and the final answer is twice that.