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Double integral under a cone

  1. Jun 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Use polar coordinates to find the volume of the given solid:
    Under the cone z = Sqrt[x^2 + y^2]
    Above the disk x^2 + y^2 <= 4

    2. The attempt at a solution
    I tried using formatting but I couldnt get it right so I'll explain...I changed variables by making the upper and lower limit of the inner integral [-2,2], with the outer integral [0,2pi]. The inner integral became integral of r^2 because Sqrt[ x^2 + y^2 ] is r then multiply by the r in r dr d(theta)... So I got

    [r^3/3] from [-2,2] which gave me 16/3. I then integrated with respect to theta from 0 to 2pi (is this correct? Is the disk around the origin??) and that gave me 32pi/3 but the answer was the original 16pi/3. This is why I think it should be integrated from 0 to pi but I cant see why because the disk lies in all 4 quadrants.
  2. jcsd
  3. Jun 11, 2010 #2


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    To cover a disk of radius 2 in polar coordinates, r does not go from -2 to 2. By convention in polar coordinates you usually take r nonnegative. Try r from 0 to 2 and theta from 0 to 2pi.
  4. Jun 11, 2010 #3
    Ah I see, looks like I may need to do work on polar coordinates :/
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