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Math Help - Question involving polar coordinates..

  1. #1
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    Question involving polar coordinates..

    Find the area of the region that is outside the graph of the first equation nd inside the graph of the second equation.
    1. r=2
    2. r^2= 8sin2x


    (x=theta)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by clev88 View Post
    Find the area of the region that is outside the graph of the first equation nd inside the graph of the second equation.
    1. r=2
    2. r^2= 8sin2x


    (x=theta)
    First, find the intersection between the circle and lemniscate:

    They intersect when (2)^2=8\sin(2\theta)\implies \tfrac{1}{2}=\sin\left(2x\right).

    Thus, 2\theta=\arcsin\left(\tfrac{1}{2}\right)=\frac{\pi  }{6},\frac{5\pi}{6}\implies \theta=\frac{\pi}{12},\frac{5\pi}{12}

    Thus, it follows that A={\color{red}2}\left[\tfrac{1}{2}\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}  } 8\sin\left(2\theta\right)-4\,d\theta\right] (Why twice the integral? I leave that for you to figure out)

    Can you continue?
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    Thumbs up

    Ya thanks!
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  4. #4
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    Hello, clev88!

    Find the area of the region that is outside r \:=\:2 and inside r^2\:=\: 8\sin2\theta
    r = 2 is a circle with radius 2.
    r^2 = 8\sin2\theta is a lemniscate ("figure 8").

    The graph looks like this . . .
    Code:
                    |
                 ...*...
              *:::::|:::::*
            *:::::::|:::::::*
           *::::::::|::::::::*
           :::::::::|:::::::::
          *:::::::::|:::::::::*
          *:::::::::|:::::::::*
          *:::::::o o o:::::::*
          *:::o     |     o:::* 
           :o       |       o:
           o        |        o
            *       |       *
          o   *     |     *   o
      - - o - - - - * - - - - o - -
          o   *     |     *   o
            *       |       * 
           o        |        o
           :o       |       o:
          *:::o     |     o:::*
          *:::::::o o o::::::::*
          *:::::::::|:::::::::*
          *:::::::::|:::::::::*
           :::::::::|:::::::::
           *::::::::|::::::::*
            *:::::::|:::::::* 
              *:::::|:::::*
                    *
                    |

    Due to the symmetry, we can find the area in quandrant 1 and multiply by 4.


    Intersection: . \begin{Bmatrix} r^2 \:=\:4 \\ r^2 \:=\:8\sin2\theta \end{Bmatrix}
    . . 8\sin2\theta \:=\:4 \quad\Rightarrow\quad \sin2\theta \:=\:\frac{1}{2} \quad\Rightarrow\quad 2\theta \:=\:\frac{\pi}{6} \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{12}


    Then: . A \;=\;4 \times \tfrac{1}{2}\int^{\frac{\pi}{12}}_0\bigg[8\sin2\theta - 4\bigg]\,d\theta

    Got it?

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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by clev88 View Post
    Find the area of the region that is outside the graph of the first equation nd inside the graph of the second equation.
    1. r=2
    2. r^2= 8sin2x

    (x=theta)
    Also see this picture

    Last edited by DeMath; July 30th 2009 at 07:52 AM.
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