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Math Help - area with polar coodinates

  1. #1
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    area with polar coodinates

    I need the area inside r=18sin(theta) but outside r=2

    Im having trouble finding the interval on which to integrate.
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  2. #2
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    Lexington, MA (USA)
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    Hello, casesam!

    Find the area inside r\:=\:18\sin\theta, but outside r\:=\:2
    A sketch might help . . .
    Code:
                  18|
                  * * *
              *:::::|:::::*
            *:::::::|:::::::*
           *::::::::|::::::::*
           :::::::::|:::::::::
          *:::::::::|:::::::::*
          *:::::::::|:::::::::*
          *:::::::::|:::::::::*
            :::::::ooo:::::::
           *:::o    |    o:::*
            *:o     |     o:*
              *     |     *
        -----o----*-*-*----o-----
                    |      2
              o     |     o
                o   |   o
                   ooo
                    |

    The graphs intersect where: . 18\sin\theta \:=\:2 \quad\Rightarrow\quad \sin\theta \:=\:\tfrac{1}{9}

    Hence: . \theta \;=\;\sin^{-1}(\tfrac{1}{9}) \;\approx\;\begin{Bmatrix}0.11134 \\ 3.03025 \end{Bmatrix}


    Due to the symmetry, we can integrate from 0.11134 to \tfrac{\pi}{2} . . . and double.

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  3. #3
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    Thanks for the graph! so now for the integration

    (2)(1/2)(integral from 0.11134 to (pi/2)) (18sin(theta))^2-(2)^2

    =(1)(integral) 324(sin(theta))^2-4
    =(4)(integral) 81(sin(theta))^2-1
    =(4)(integral)81((1/2)(1-cos(2theta)))-1
    =(4)(integral) (81/2)-(81/2)cos(2theta)-1
    =4[(79/2)theta-(81/4)sin(2theta)] integrated from 0.11134 to (pi/2)

    Is this right? and what final answer do you get with many decimal places please.

    Thanks in advance
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