# Triple intergrals exer

1. Sep 13, 2004

### ori

how can i calculate this integral value :
SSSdxdydz/(x^2+y^2+z^2)
at the area: x^2+y^2+(z-1)^2<=1

thats what i tried - i got double integral that i dont know how to solve
i tried to use the ball cordinations
x=rcos(t)sin(p)
y=rsin(t)sin(p)
z=rcos(p)+1
j=rsin^2(p)
0<=r<=1
0<=t<2pi
0<=p<=pi

i get:

SSS drdtdp/[r^2cos^2(t)sin^2(p)+r^2sin^2(t)sin^2(p)+(rcos(p)+1)^2]
0<=r<=1
0<=t<2pi
0<=p<=pi

SSS drdtdp/[r^2sin^2(p)+r^2cos^2(p)+2rcos(p)+1]

2pi * SS drdp/[r^2+2rcos(p)+1]

what's next?

10x

2. Sep 13, 2004

### Feynman

You have an error :
the spherical coordinate is:
x=rcos(t)sin(p)
y=rsin(t)sin(p)
z=rcos(p)
only!!!!!!!

3. Sep 13, 2004

### ori

my transformation deals with the area that the integral works on
urs deals with the integrand , both ways i dont know how to solve the integral
that we get