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Math Help - Double integral in polar coordinates- funky bounds!?

  1. #1
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    Double integral in polar coordinates- funky bounds!?

    Hello,

    The problem is to convert a cartesian double integral to polar coordinates and evaluate. I'm fine with evaluating, I just require some assistance determining my bounds in polar coordinates

    double integral f(x,y) = y and order of integration is dy dx
    1< x< 2
    0 < y < sqrt{4-(x^2)}
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  2. #2
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    Have you tried making a sketch of this region?
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    first of all, you need to determine the limits of your angle θ. since the region of integration is bounded below by the x-axis, one limit for θ is θ = 0.

    the second limit will be the angle of the line through the origin and the point where the vertical line x = 1, and the semi-circle y = √(4 - x^2) intersect.

    this should not be all that hard for you to deduce (it is a commonly encountered angle). the limits of r will depend on θ, you have two curves:

    r = h1(θ), the vertical line x = 1, this will be the hard one to write down, and

    r = h2(θ), the semi-circle y = √(4 - x^2). this should be obvious.

    writing down f in polar coordinates will not be difficult, but don't forget the adjustment factor for dy dx.
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    Quote Originally Posted by Prove It View Post
    Have you tried making a sketch of this region?
    I made a sketch, it looks like the line x=1 makes the region into roughly half of the quarter circle in quadrant one. I know how to find the angles using trig for d\theta , but I cannot figure out the upper and lower radius limits
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  5. #5
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    what is the equation for a circle in polar coordinates? isn't it a really, really simple one?

    for example, isn't the circle x^2 + y^2 = 1 just the constant function r = 1 in polar coordinates?

    the equation for the vertical line is more challenging. see if you can put the fact that x = rcosθ to good use, if x is constant.
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  6. #6
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    What is the radius when x = 1 (It will be in terms of \theta)? What is the radius when x = 2?
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