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Math Help - Area between two circles

  1. #1
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    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
    Area between two circles-codecogseqn.gif
    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?
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  2. #2
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    Re: Area between two circles

    Area between two circles-17-nov-13.png
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  3. #3
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    Re: Area between two circles

    Hello, nubshat!

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

    The graph looks like this . . .

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

    and A can be found with: . \int^2_1\sqrt{4-x^2}\,dx
    Thanks from nubshat
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  4. #4
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    Re: Area between two circles

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  5. #5
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    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 \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 (120/360)\pi r^2, and the area of the triangle POR is (1/2)PO\cdot OR\sin\angle POR =\sqrt{3}. From here, the area of PLRQ is (4/3)\pi-\sqrt{3}, and the final answer is twice that.
    Thanks from nubshat
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