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Math Help - Polar coordinates area

  1. #1
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    Polar coordinates area

    r=cos(t) r=sin(t)

    find the area of intersection.

    I come up with integrals of cos^2 t - sin^2 t from 0 to pi/4, and from 5pi/4 to 2pi, sin^2 t - cos^2 t from pi/4 to 5pi/4.

    is that correct and, is there a simpler way to tackle the problem?

    Thank you very much!
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  2. #2
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    Quote Originally Posted by kyva1929 View Post
    r=cos(t) r=sin(t)

    find the area of intersection.

    I come up with integrals of cos^2 t - sin^2 t from 0 to pi/4, and from 5pi/4 to 2pi, sin^2 t - cos^2 t from pi/4 to 5pi/4.

    is that correct and, is there a simpler way to tackle the problem?

    Thank you very much!
    From the graph below, we should have

    Area= 2*\frac{1}{2}\int_{0}^{\frac{\pi}{4}}sin^{2}(\thet  a)d\theta
    Attached Thumbnails Attached Thumbnails Polar coordinates area-two-circles.gif  
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  3. #3
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    Quote Originally Posted by curvature View Post
    From the graph below, we should have

    Area= 2*\frac{1}{2}\int_{0}^{\frac{\pi}{4}}sin^{2}(\thet  a)d\theta
    Oh! I just figure it out by plotting the graph with maple.

    What do you use to plot the graph attached? It seems nicer than the one I plotted with maple.

    Thank you!
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  4. #4
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    Quote Originally Posted by kyva1929 View Post
    r=cos(t) r=sin(t)

    find the area of intersection.

    I come up with integrals of cos^2 t - sin^2 t from 0 to pi/4, and from 5pi/4 to 2pi, sin^2 t - cos^2 t from pi/4 to 5pi/4.

    is that correct and, is there a simpler way to tackle the problem?

    Thank you very much!
    \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2{t} \, dt

    or

    \int_0^{\frac{\pi}{4}} \sin^2{t} \, dt
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  5. #5
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    Quote Originally Posted by kyva1929 View Post
    Oh! I just figure it out by plotting the graph with maple.

    What do you use to plot the graph attached? It seems nicer than the one I plotted with maple.

    Thank you!
    The graph was plotted with maple.
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