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Math Help - Polar Coor Intergration Help Part Duex

  1. #1
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    Polar Coor Intergration Help Part Duex

    Hey again guys
    for some reason I cant get this to work again.

    Evaluate the integral by changing to polar coordinates.
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  2. #2
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    Quote Originally Posted by TheUnfocusedOne View Post
    Hey again guys
    for some reason I cant get this to work again.

    Evaluate the integral by changing to polar coordinates.
    First step, draw a picture. x lies between 0 and 1/2 and, for each x, y is between the straight line y= \sqrt{3}x and the unit circle x^2+ y^2= 1.

    In polar coordinates, the unit circle is given, of course, by r= 1 and, since the slope of a line is the tangent of the angle the line makes with the x-axis, the line y= \sqrt{3} x makes an angle [tex]\pi/3[tex]. Further, that line just happens to cross the circle where x^2+ y^2= x^2+ (\sqrt{3} x)^2= 4x^2= 1: x= 1/2.

    The region in this integral s the segment of the unit circle in which r goest from 0 to 1 and \theta goes from \pi/3 to \pi/2.
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  3. #3
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    r goes from 0 to 1?
    you gotta be kidding me....
    thanks a lot
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