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Double Integral over circle

  1. Jun 10, 2010 #1
    So I have to use the type I type II region formula to find the volume under the equation (2x-y) and over the circular domain with center (0,0) and radius 2. Do I have to split this circle into hemispheres and treat it as 2 type I domains? I got the following limits for the top half, but I get stuck when integrating:

    y limits:
    Upper: Sqrt(2 - x^2) from the equation 2 = y^2 + x^2
    Lower: 0

    X limits:
    Upper: 2
    Lower: -2

    So I have to find the integral with respect to y of 2x-y with limits 0 to Sqrt[2-x^2]

    After integrating with respect to Y I got:

    2x(Sqrt[2-x^2]) - 1 + (x^2)/2

    Is this correct to start with? Then integrate with respect to x from -2 to 2?
  2. jcsd
  3. Jun 10, 2010 #2


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    Hi Shaybay92! :smile:

    (have a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)

    Can you please clarify the question? :confused:

    Are you trying to find a 2D area, or a 3D volume?

    By "the equation (2x-y)" do you mean the line (or plane) with 2x-y = 0?

    If so, isn't this just the area of a semi-circle (or the volume of a hemisphere)?

    (I'm not familiar with the "type and I type II" classification, but it looks like you need to use the area or volume under 2x-y=0 separately from the area or volume of the cap that's "clear" of 2x-y=0)
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