1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Triple Integral Problem

  1. May 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the triple integrals [tex]\oint\oint\oint_{W}{f(x,y,z)dV[/tex]:

    [tex]e^{x^{2}+y^{2}+z}, (x^{2}+y^{2}) \leq z \leq {(x^{2}+y^{2}})^{1/2}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    So I know I need to probably switch to cylindrical coordinates. But I'm getting confused about the limits of integration. The way that I see it, since there's no limits of integration for z then the volume which is the between the two parabolas goes to infiniti? But since it's the same above and below the z=0 plane then doesn't that just come out to zero? I dunno I guess if I do the integral from x going from -infiniti to infiniti and use an improper integral from some (a = infiniti) with the limits of integration for z being then from a to a which is obviously zero. I don't think that he would make up a problem like that though. Am I completely wrong in my thinking. THANKS!
    Last edited: May 10, 2008
  2. jcsd
  3. May 10, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    You do NOT have two paraboloids- and none of the figure is below the z= 0 plane. [itex]z= x^2+ y^2 [/itex] is a paraboloid with vertex at (0,0,0), axis the z-axis, and opening upward. [itex]z= (x^2+ y^2)^{1/2}[/itex] is the upper half of the cone z^2= x^2+ y^2[/itex]. It does not go to infinity. The two surfaces intersect at [itex]z= x^2+ y^2= (x^2+ y^2)^{1/2}[/itex]. Putting that into cylindrical coordinates makes it particularly easy: [itex]z= r^2[/itex] and [itex]z= r[/itex] interxect when [itex]r^2= r[/itex].
  4. May 11, 2008 #3
    Ok So know when I try to do the integration am I correct to use the limits of integration of that z goes from 0 to 1 and theta goes from 0 to 2pi and then r goes from root z to z? Also I tried using z from r to r^2 r from 0 to 1 and theta from 0 to 2pi, and they are all giving me strange integrals. Are these limits correct? I'm having a hard time figuring out how to change the limits of integration when switching to cylindrical, especially with regards to r.
  5. May 11, 2008 #4
    I'm still having trouble figuring out the limits of integration on this one. Every way that I do it I keep having to take the integral of

    x*e^(x^2+x) which as far as I know isn't possible to do. I tried plugging it into a numerical solver and it gave me an exact answer that looked like some sort of estimation. HELP! I have a mid-term Tues. and this determing the limits of integration is gonna screw me big time.
  6. May 11, 2008 #5
    I'm begining to think that my teacher made an error in writing this problem. It appears to be unsolvable through all my efforts.
  7. May 11, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    No, z does NOT go from 0 to 1. The whole point of what you are doing is that z goes from the lower of those two surfaces (z= x2+ y2) to the higher surface ([itex]z= \sqrt{x^2+ y^2}[/itex]). Since I have already told you that the two surfaces intersect where r2= r- which tells you r= 0 or r= 1, projected down into the z= 0 plane, the two surfaces project the the area from (0,0) to the circle about (0,0) with radius 1. That is, [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex] while r goes from 0 to 1.

    [tex]\int_{\theta= 0}^{2\pi} \int_{r=0}^1 \int_{z=r^2}^r e^{r^2+ z} rdzdrd\theta[/itex]
    I think that's the integral you need to do. You never did say that [itex]f(x,y,z)= e^{x^2+y^2+ z}[/itex].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Triple Integral Problem
  1. Triple Integral problem (Replies: 11)