Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    HallsofIvy

    User Avatar
    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

    HallsofIvy

    User Avatar
    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].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook