Answer to 17 is 4pi /5

But I keep getting 8pi / 5

Helps

do you convert the last 2 integrals to polar?
if so, what are your bounds after you convert to polar?

E: also, be sure to add in the "r" when changing to polar (r dr d(theta))

This post was edited by TritonV8 on Apr 29 2013 07:29pm

the simplest way is to convert to cylinder coordinates.

z=z
rho or sometimes called r = sqrt(x^2+y^2)
since x^2+y^2 =1
rho is just between 0 and 1
phi is between 0 and 2pi since we are integrating over the whole circle.


so we have the integral p^4 dpdphidz , 0<p<1 , 0<phi<2pi, -1<z<1
because (x^2+y^2)^(3/2)=(p^2)^(3/2) and you also must multiply this by p to transform from dxdydz to dpdphidz

integrating dp we get from p =0 to 1
1/5dphidz
integrating dphi we get from phi from 0 to 2pi
2pi/5 dz
integrating z from -1 to 1
4pi/5

i assumed your familiar with change of variables with integrals. hope that helps ill have more time later to explain more in-depth if you need