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Math Help - Finding area bounded by two cardioids

  1. #1
    s7b
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    Finding area bounded by two cardioids

    Find the region common to the interiors of the cardioids r=1+costheta and r=1-costheta


    -I understand what the picture looks like I'm having difficulty with the limits of integration. Since both the circles are the same you can find the area of the top one and multiply the double integral by 2. in this case I'm just not sure what the limits of integration of r would be. I think is would be from zero to ?? and also I think the limit of integration for theta would be 0 to pi.

    Help!

    One more thing. Is the actual integral just of rdrd(theta)
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  2. #2
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    Hello, s7b!

    Are you sure you have the right graph?


    Find the area of the region common to the interiors of the cardioids;
    . . r\:=\:1+\cos\theta\,\text{ and }\,r\:=\:1-\cos\theta
    Code:
                              |
                    *         |         *
              *           *   |   *           *
                              *
          *                 *:|:*                 *
                           *::|::* 
        *                   *:|:*                   *
    - - * - - - - - - - - - - * - - - - - - - - - - * - - 
        *                   *:|:*                   *
                           *::|::*
          *                 *:|:*                 *
                              *
              *           *   |   *           *
                    *         |         *
                              |

    Then intersect when: . 1 + \cos\theta \:=\:1-\cos\theta \quad\Rightarrow\quad \theta \:=\:\pm\tfrac{\pi}{2}

    The region has four-way symmetry.
    We can find the area in Quadrant 1 and multiply by 4.

    The integral is: / \text{Area} \;=\;4 \times\tfrac{1}{2}\int^{\frac{\pi}{2}}_0\left(1 - \cos\theta\right)^2d\theta

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  3. #3
    MHF Contributor
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    common area using symmetry ...

    A = 4 \int_0^{\frac{\pi}{2}} \frac{(1 - \cos{\theta})^2}{2} \, d\theta

    or ...

    A = 4 \int_{\frac{\pi}{2}}^{\pi} \frac{(1 + \cos{\theta})^2}{2} \, d\theta
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