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Math Help - Area in polar coordinates

  1. #1
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    Area in polar coordinates

    Calculate in polar coordinates the area of the region outside of P=1, P=2cos(\theta) and in inside P=2

    This is correct:

    A=2\int_{0}^{\frac{\pi}{3}} \int_{2cos\theta}^{2} PdPd\theta + 2\int_{\frac{\pi}{3}}^{\pi} \int_{1}^{2} PdPd\theta

    and the answer is: (\frac{8\pi}{3}-\frac{\sqrt{3}}{2})
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  2. #2
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    Hello, Apprentice123!

    What does P stand for in your equation?


    Calculate in polar coordinates the area of the region
    outside of r_1=1,\;\;r_2 =2\cos\theta, and inside r_3=2
    I don't see the need for double integrals . . .

    . . \begin{array}{c}r_1\text{ is a circle: center (0,0), radius 1} \\ r_2 \text{ is a circle: center (1,0), radius 1} \\ r_3\text{ is a circle: center (0,0), radius 2} \end{array}

    The region is symmetric to the x-axis.
    . . We can determine the area of the upper half and multiply by 2.


    From \theta = 0 to \theta = \tfrac{\pi}{3}, the area is between r_3\text{ and }r_2

    . . A_1 \;=\;\tfrac{1}{2}\int^{\frac{\pi}{3}}_0\bigg[ 2^2 - (2\cos\theta)^2\bigg]\,d\theta


    From \theta = \tfrac{\pi}{3} to \theta = \pi, the area is between r_3\text{ and }r_1

    . . A_2 \;=\;\tfrac{1}{2}\int^{\pi}_{\frac{\pi}{3}} \bigg[2^2 - 1^2\bigg]\,d\theta


    Add the two areas and multiply by 2 . . .

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