Triple Integral in Cylindrical Coordinates

  • Thread starter daveyman
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  • #1
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Revised question is below.
 
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  • #2
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Homework Statement


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.


Homework Equations


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


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?
 
  • #3
gabbagabbahey
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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]
 
  • #4
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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.
 
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  • #5
gabbagabbahey
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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.
 
  • #6
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Oh I get it. Thank you so much!
 

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