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Why is volume generated 0?

  1. Mar 23, 2010 #1
    Evaluate [tex]\int\int\int y dV[/tex] where the solid generated lies between the cylinders x2+y2=1 and x2+y2=4, above the xy-plane and below the plane z=x+2.

    I wrote out the integral [tex]\int\int r*r sin \theta dz dr d\theta[/tex] and z is integrated from x+2 to 0, r integrated from 2 to 1 and [tex]\theta[/tex] from [tex]2\pi[/tex] to 0. But I ended up with [tex]\frac{-7}{3}(x+2)cos(2\pi - 0)[/tex] which gives me 0?? I tried visualizing the volume generated and its like a donut with the top cut away at an angle. So how is the volume 0? Did I do something wrong?
     
    Last edited: Mar 23, 2010
  2. jcsd
  3. Mar 23, 2010 #2

    Mark44

    Staff: Mentor

    Your integral doesn't represent volume, so it's possible that its value is 0. If the integrand were 1 instead of 1, then it would represent volume.
     
  4. Mar 23, 2010 #3
    You mean the one I wrote or the equation given?
     
  5. Mar 24, 2010 #4
    *bump*
     
  6. Mar 24, 2010 #5

    Pengwuino

    User Avatar
    Gold Member

    Please do not bump threads.

    The integral of JUST dV would represent a volume. However, the integration of a function IN a volume no longer represents the volume.

    For example,

    [tex]\int\limits_0^L {dx} [/tex]

    gives you the length of something from 0-> L, which is just a length of L. However,

    [tex]\int\limits_0^L {x^2 dx} [/tex]

    no longer gives a length.
     
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