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Math Help - Areas under and between polar curves

  1. #1
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    Areas under and between polar curves

    Find the area of the region outside r=9+9sin(theta), but inside r=27sin(theta)
    I did the integral of .5((27sin(theta))^2-(9+9sin(theta))^2), from pi/6 to 5pi/6, but got the wrong answer. Any hints?

    Find the area inside the curve r^2 = 72cos(2theta)
    Do I just take the square root of the entire equation? If so, how do I integrate the square root of cosine?

    Find the area inside one loop of the curve r^2=4sin(2theta)
    Same question as the previous one
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  2. #2
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    Quote Originally Posted by mistykz View Post
    Find the area of the region outside r=9+9sin(theta), but inside r=27sin(theta)
    I did the integral of .5((27sin(theta))^2-(9+9sin(theta))^2), from pi/6 to 5pi/6, but got the wrong answer. Any hints?
    Did you get 81{\pi}?.


    Find the area inside one loop of the curve r^2=4sin(2theta)
    Same question as the previous one
    When you set up the polar form of integration, the square eliminates the radical.

    \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\sqrt{4si  n(2{\theta})}\right)^{2}d{\theta}=2\int_{0}^{\frac  {\pi}{2}}sin(2{\theta})d{\theta}
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    Hello, mistykz!

    You seem to be forgetting the area formula: . A \;=\;\frac{1}{2}\int^{\beta}_{\alpha} r^2\,d\theta


    Find the area inside the curve: . r^2 \:= \:72\cos(2\theta)

    Do I just take the square root of the entire equation?
    If so, how do I integrate the square root of cosine?
    Look at the formula again!
    It says: . A \;=\;\frac{1}{2}\int^{\beta}_{\alpha}{\color{red}r  ^2}\,d\theta

    So we have: . A \;=\;\frac{1}{2}\int^{\beta}_{\alpha}72\cos2\theta  \,d\theta \;=\;36\int^{\frac{\pi}{4}}_{\text{-}\frac{\pi}{4}} \cos2\theta\,d\theta


    Got it?

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