Triple Integral for Cylindrical Coordinates in a Parabolic Region

[robbondo]

Homework Statement

Find the triple integrals \oint\oint\oint_{W}{f(x,y,z)dV:

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

Homework Equations

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 don't know 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!

 

[HallsofIvy]

You do NOT have two paraboloids- and none of the figure is below the z= 0 plane. z= x^2+ y^2 is a paraboloid with vertex at (0,0,0), axis the z-axis, and opening upward. z= (x^2+ y^2)^{1/2} 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 z= x^2+ y^2= (x^2+ y^2)^{1/2}. Putting that into cylindrical coordinates makes it particularly easy: z= r^2 and z= r interxect when r^2= r.

 

[robbondo]

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.

 

[robbondo]

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 going to screw me big time.

 

[robbondo]

I'm beginning to think that my teacher made an error in writing this problem. It appears to be unsolvable through all my efforts.

 

[HallsofIvy]

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 (z= \sqrt{x^2+ y^2}). 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, \theta goes from 0 to 2\pi while r goes from 0 to 1.

\int_{\theta= 0}^{2\pi} \int_{r=0}^1 \int_{z=r^2}^r e^{r^2+ z} rdzdrd\theta[/itex]<br /> I <b>think</b> that&#039;s the integral you need to do. You never <b>did</b> say that f(x,y,z)= e^{x^2+y^2+ z}.