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Math Help - Integral problem

  1. #1
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    Integral problem

    Hi, I am having trouble with this problem. Any help would be appreciated.

    Calculate the area of the region bounded by the curve

    (x^2+y^2-x)^2=x^2+y^2 and the circle x^2+y^2=sqrt(3) y

    Thanks
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  2. #2
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    Hello, 0pokerman0!

    Calculate the area of the region bounded by
    the curve: . (x^2+y^2-x)^2\:=\:x^2+y^2 .and the circle: . x^2+y^2 \:=\:\sqrt{3}y
    Convert to polar coordinates . . .


    (x^2+y^2 - x)^2 \:=\:x^2+y^2\quad\Rightarrow\quad (r^2 - r\cos\theta)^2 \:=\:r^2

    Take the square root: . r^2 - r\cos\theta \:=\:r

    Divide by r\!:\;\;r - \cos\theta \:=\:1\quad\Rightarrow\quad r \:=\:1 + \cos\theta . . . a cardoid.


    x^2 + y^2 \:=\:\sqrt{3}y\quad\Rightarrow\quad r^2 \:=\:\sqrt{3} r\sin\theta \quad\Rightarrow\quad r \:=\:\sqrt{3}\sin\theta

    . . a circle with center on the "y-axis", through the origin, with diameter \sqrt{3}

    Code:
                |
                |        *
                o  *            *
           o    *::::o
          o   *:|:::::o            *
          o  *::|:::::o
          o  *::|:::::o             *
           o  *:|::::o
          - - - o - - - - - - - - - * - -
              * |
             *  |                   *
             *  |
              * |                  *
                * 
                |  *            *
                |        *
                |


    Now try it . . .

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  3. #3
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    what would the intergrand be?
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  4. #4
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    Hello again, 0pokerman0!


    We see that the two curves intersect at the origin.

    Find the other intersection:

    . . \sqrt{3}\sin\theta \:=\:1 + \cos\theta \quad\Rightarrow\quad \sqrt{3}\sin\theta - \cos\theta \:=\:1

    Divide by 2: . \frac{\sqrt{3}}{2}\sin\theta - \frac{1}{2}\cos\theta \:=\:\frac{1}{2}


    Since: \frac{\sqrt{3}}{2} = \cos\frac{\pi}{6},\;\;\frac{1}{2} = \sin\frac{\pi}{6}

    . . we have: . \cos\frac{\pi}{6}\sin\theta - \sin\frac{\pi}{6}\cos\theta \:=\:\frac{1}{2}\quad\Rightarrow\quad\sin\left(\th  eta - \frac{\pi}{6}\right) \:=\:\frac{1}{2}

    Then: . \theta - \frac{\pi}{6} \:=\:\frac{\pi}{6}\quad\Rightarrow\quad \boxed{\theta \:=\:\frac{\pi}{3}}


    This requires two integrals . . .
    . . the area in the circle from 0\text{ to }\frac{\pi}{3}
    . . the area in the cardioid from \frac{\pi}{3}\text{ to }\pi

    A \;\;=\;\;\frac{1}{2}\int^{\frac{\pi}{3}}_0\bigg[\sqrt{3}\sin\theta\bigg]^2d\theta \;+ \;\frac{1}{2}\int^{\pi}_{\frac{\pi}{3}}\bigg[1 + \cos\theta\bigg]^2d\theta

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