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Math Help - Polar coordinates area of the region bounded problem

  1. #1
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    Polar coordinates area of the region bounded problem

    Find the area of the region bounded by: r = 9 − 4 sinθ

    I tried this multiple times and the answer i got was 381.7035 but that is not correct.
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  2. #2
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    Quote Originally Posted by muddyjch View Post
    Find the area of the region bounded by: r = 9 − 4 sinθ

    I tried this multiple times and the answer i got was 381.7035 but that is not correct.
    It looks straightforward to me (and I do NOT get 381.7035). What exactly did you do?
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  3. #3
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    Hello, muddyjch!

    Find the area of the region bounded by: . r \:=\: 9 - 4\sin\theta

    This is a heart-shaped region symmtetric to the "y-axis".

    We can integrate from \text{-}\frac{\pi}{2} to +\frac{\pi}{2} and multiply by 2.


    We have: . \displaystyle{A \;=\;2 \times \frac{1}{2}\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} (9 - 4\sin\theta)^2\,d\theta}

    . . . . . . . . . . \displaystyle{=\;\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} \left(81 - 72\sin\theta + 16\sin^2\!\theta\right)\,d\theta }

    . . . . . . . . . . \displaystyle{=\;\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}\bigg(81 - 72\sin\theta + 16\left[\tfrac{1-\cos2\theta}{2}\right]\bigg)\,d\theta  }

    . . . . . . . . . . \displaystyle{=\; \int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}\bigg(81 - 72\sin\theta + 8 - 8\cos2\theta\bigg)\,d\theta }

    . . . . . . . . . . \displaystyle{=\;\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}\bigg(89 - 72\sin\theta - 8\cos2\theta\bigg)\,d\theta }

    . . . . . . . . . . \displaystyle{=\;89\theta + 72\cos\theta - 4\sin2\theta\,\bigg]^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} }

    . . . . . . . . . . =\; \bigg[89\left(\frac{\pi}{2}\right) + \cos\frac{\pi}{2} - 4\sin\pi\bigg] - \bigg[89\left(\text{-}\frac{\pi}{2}\right) + 72\cos\left(\text{-}\frac{\pi}{2}\right) - 4\sin(\text{-}\pi)\bigg]

    . . . . . . . . . . =\;\bigg(\frac{89\pi}{2} + 0 - 0\bigg) - \bigg(\text{-}\frac{89\pi}{2} + 0 - 0\bigg)

    . . . . . . . . . . =\;89\pi \;\approx\;279.6017462


    But check my work . . . please!
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