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Triple Integration of a Strange Cylinder

  1. Feb 8, 2009 #1
    Hey all,

    If anyone has some hints on how to do this one it would be much appreciated:

    Find the volume of the region given by x2 + y2 ≤ a2, 0 ≤ z ≤ x.

    So I've gone to cylindrical polars, and threw in the Jacobian, r. If I integrate with my bounds being:

    0 to a, 0 to pi, 0 to rcos(theta) of r dz d(theta) dr I'm getting 0 which can't be. What is it that I'm missing here. Any hint would be great.
    Last edited: Feb 9, 2009
  2. jcsd
  3. Feb 9, 2009 #2


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    Science Advisor

    So you are integrating
    [tex]\int_{r=0}^a\int_{\theta= 0}^\pi\int_{z= 0}^{r cos(\theta)} r dzd\theta dr[/itex]?

    If you look closely at this situation you will see that this is symmetric about the x axis and so, with z= x as a boundary, half the volume is below the z= 0 plane, half above. Since your integral treats the volume below the z= 0 plane as negative, those cancel. The true "volume" is twice the volume above the z-axis. You get the volume above the z= 0 plane by assuring that x is positive: by taking [itex]\theta[/itex] from [itex]-\pi/2[/itex] to [itex]\pi/2[/itex].
  4. Feb 9, 2009 #3
    Thank you very much. That's just the bit I needed. Sorry about not using the Latex code. I'm new to this site.
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