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Volume using double integral and polar coordinates

  1. Apr 26, 2010 #1
    Find the volume under the cone z = sqrt ( x2+y2 ) and on the disk x2+y2 < 4. Use polar coordinates.

    Graphing x2+y2 < 4, I get a circle centered at 0,0 with radius of 2
    So theta goes from 0 to 2pi

    Also, since x2+y2 < 4
    This means that r^2 < 4
    so -2 < r < 2

    [tex]\int_{0}^{2\pi}\int_{-2}^{2}r\sqrt{x^{2}+y^{2}} dr d\theta=\int_{0}^{2\pi}\int_{-2}^{2}r^{2} dr d\theta
    = \frac{32\pi}{3}[/tex]
    but the answer is 16pi/3

    Can anyone help me
  2. jcsd
  3. Apr 26, 2010 #2


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    Homework Helper

    Your integration boundaries for r are not correct.
  4. Apr 26, 2010 #3


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    Gold Member

    radius can't be negative, its length is still 2
    [tex]\int_{0}^{2\pi}\int_{0}^{2}r^{2} dr d\theta [/tex]

    (although if you negate the radius, you increase theta by pi):
    [tex]\int_{pi}^{2\pi}\int_{-2}^{2}r^{2} dr d\theta [/tex]

    (probably just easier to think of radius as a non-negative variable)
    Last edited: Apr 26, 2010
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