Question

Evaluate the iterated integral:

http://www.wolframalpha.com/input/?i=integrate+e%5E-p%5E3+p%5E2+from+phi%3D0+to+pi%2F2%2Ctheta%3D0+to+pi%2Cp%3D0+to+2

Answer 1

The mistake I made in this problem was in evaluating the first integral:

I'm confused on how they get their answer
http://www.wolframalpha.com/input/?i=integrate+e%5E-p%5E3+p%5E2+from+p%3D0+to+2
I'll post up qhat I have

Answer 2

\[\int\limits_{0}^{pi/2}\int\limits_{0}^{pi}\int\limits_{0}^{2}e^-p^3p^2 dp d \theta d \phi\]
\[=\frac{e^-p^3p^3}{3} d \theta d \phi\]

Answer 3

something that might help, notice how p is not a function of theta or phi, that all three variables are independent? that means you can actually seperate the three integrals like this:
\[\int\limits_{0}^{pi/2} d \phi \int\limits_{0}^{pi} d \theta \int\limits_{0}^{2} e^{-p ^{3}} dp\]

Answer 4

Yeah, someone's mentioned that to me before, I always seem to forget that...

Answer 5

so you would get :
\[(pi/2)(pi) \int\limits_{0}^{2} e ^{-p ^{3}}dp\]

Answer 6

alright, i'll do this out now and see what I get.

Answer 7

and to help with the integral of e to the (ugly) you should understand how chain rule works involving e^(f(x))?
basically if i have the derivative of e^(f(x)) then the derivative will be [e^(f(x))]* f'(x)

Answer 8

yeah I get that.

Answer 9

^ you forgot p^2.

Answer 10

well, 3p^2 is f'(x) which i figured you could do yourself

Answer 11

srry -3p^2

Answer 12

alright