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Homework Help: Triple Integral in Cylindrical Coordinates

  1. Nov 9, 2008 #1
    Revised question is below.
    Last edited: Nov 9, 2008
  2. jcsd
  3. Nov 9, 2008 #2
    1. The problem statement, all variables and given/known data
    Evaluate [tex]\int\int\int_E{\sqrt{x^2+y^2} dV[/tex] where E is the region that lies inside the cylinder [tex]x^2+y^2=16[/tex] and between the planes z=-5 and z=4.

    2. Relevant equations
    For cylindrical coordinates, [tex]r^2=x^2+y^2[/tex].

    3. The attempt at a solution
    The inside of the integral becomes [tex]\sqrt{r^2}=r[/tex]. Then I integrated using the following bounds: [tex]\int _0^{2*\pi }\int _{-5}^4\int _0^4r drdzd\theta[/tex]

    However, this gives me an answer of [tex]144\pi[/tex]. I've tried several things in Mathematica and I finally tried [tex]\int _0^{2*\pi }\int _{-5}^4\int _0^4r^2drdzd\theta[/tex] which actually gave me the right answer of [tex]384\pi[/tex]. However, integrating over [tex]r^2[/tex] makes no sense.

    Any ideas?
  4. Nov 9, 2008 #3


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    The infinitesimal volume element in cylindrical coordinates is [itex]dV=rdrd\theta dz[/itex] not just [itex]drd\theta dz[/itex]
  5. Nov 9, 2008 #4
    Thank you for your quick response! I have one more question, though.

    Doesn't the [tex]dr[/tex] take care of the radial component of the volume? Why does it need to be [tex]r dr[/tex]?

    I realize this is a naive question, but I really appreciate your help.
    Last edited: Nov 9, 2008
  6. Nov 9, 2008 #5


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    [itex]dr[/itex] does take care of the radial component, but the tangential component is [itex]r d\theta[/itex] not [itex]d \theta[/itex]...remember that the arc length subtended by an angle [itex] \theta[/itex] is [itex]r \theta[/itex]...the same is true for the infinitesimal change in arc length.
  7. Nov 9, 2008 #6
    Oh I get it. Thank you so much!
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