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  1. #1
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    calc help

    Using polar coordinates evaluate the integral which gives the area which lies in the first quadrant between the circles x^2+y^2=144 and x^2-12x+y^2=0.
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  2. #2
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    Hello, karnold9!

    Using polar coordinates evaluate the integral which gives the area which lies
    in the first quadrant between the circles x^2+y^2\:=\:144\text{ and }x^2-12x+y^2\:=\:0
    Convert the equations to polar form:

    . \begin{array}{cccc}x^2 + y^2\:=\:144 & \Rightarrow & r \:=\:12 & \text{Circle: center (0,0), radius 12}\\<br />
x^2-12x + y^2 \:=\:0 & \Rightarrow & r \:=\:12\cos\theta & \text{Circle: center (6,0), radius 6}\end{array}


    A \;=\;\frac{1}{2}\int^{\frac{\pi}{2}}_0\left[12^2 - (12\cos\theta)^2\right] \,d\theta \;=\;72\int^{\frac{\pi}{2}}_0\left[1 - \cos^2\!\theta\right]\,d\theta \;=\;72\int^{\frac{\pi}{2}}_0\sin^2\theta\,d\theta<br />

    . . Got it?

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