# Tripple integral

1. Jul 10, 2008

### jaredmt

1. The problem statement, all variables and given/known data
Using cylindrical coordinates, evaluate ʃʃʃe^z where E is enclosed by the paraboloid z = 1 + x^2 + y^2 and the cylinder x^2 + y^2 = 5 and the xy-plane.

2. Relevant equations

3. The attempt at a solution
here are the integrals in xyz coordinates:
0>z>1 + x^2 + y^2
-(5-x^2)^.5 > y > (5 - x^2)^.5
-(5)^.5 > x > (5)^.5

i double checked that with my proffessor so it has to be right. but i must be doin something wrong when i switch into polar coordinates. here is what the polar limits are (i thought):
0 > z > 1 + r^2
0 > r > 5^.5
0 > ɵ > 2pi

i keep getting the wrong answer. if anything is wrong i would think it is z. unless maybe it is right and i am just integrating wrong but i checked over a million times and didnt c any mistakes :( i have:
ʃʃʃ(e^z)rdzdrdɵ

edit: let me know if u need more info. i can type out all my integration steps if u think my limits look good

Last edited: Jul 10, 2008
2. Jul 10, 2008

### Tom Mattson

Staff Emeritus

No dz integration?

3. Jul 10, 2008

### HallsofIvy

Staff Emeritus
You realize this makes no sense don't you? 1+ x2+ y2> 1 for all x, y. z can't be less than 0 and greater than 1! Surely you mean 0< z< 1+ x2+ y2.

Again 0< z< 1+ r2

So you have
[tex]\int_{r= 0}^{\sqrt{5}}\int_{\theta= 0}^{2\pi}\int_{z= 0}^{1+r^2} e^z dzd\theta dr[/itex]
Is that right?

4. Jul 10, 2008

### jaredmt

yes that is right. sorry i typed it up wrong, my bad. im not sure how to use those symbols either but w/e

does the limits look right to u?

edit: i changed it and added dz, should be typed up correctly now

edit: o sorry, actually the same thing u wrote, but multiplied by r. i thought u had to do that when u change to polar. idk maybe it is different in this case for some reason