Evaluate Tripple Integral of e^z in Cylindrical Coordinates

  • Thread starter Thread starter jaredmt
  • Start date Start date
  • Tags Tags
    Integral
jaredmt
Messages
120
Reaction score
0

Homework Statement


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.

Homework Equations


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 doing 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:
Physics news on Phys.org
jaredmt said:
ʃʃʃ(e^z)rdrdɵ


No dz integration? :confused:
 
jaredmt said:

Homework Statement


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.


Homework Equations





The Attempt at a Solution


here are the integrals in xyz coordinates:
0>z>1 + x^2 + y^2
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.

-(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 doing something wrong when i switch into polar coordinates. here is what the polar limits are (i thought):
0 > z > 1 + r^2
Again 0< z< 1+ r2

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)rdrdɵ

edit: let me know if u need more info. i can type out all my integration steps if u think my limits look good
So you have
\int_{r= 0}^{\sqrt{5}}\int_{\theta= 0}^{2\pi}\int_{z= 0}^{1+r^2} e^z dzd\theta dr[/itex]<br /> Is that right?
 
yes that is right. sorry i typed it up wrong, my bad. I am 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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top