How can I solve this triple integral using spherical coordinates?

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SUMMARY

The discussion focuses on solving the triple integral of the function SSS dxdydz/(x^2+y^2+z^2) over the region defined by x^2+y^2+(z-1)^2<=1 using spherical coordinates. The user attempted to convert the integral into spherical coordinates but encountered difficulties in solving it. Key transformations include x=rcos(t)sin(p), y=rsin(t)sin(p), and z=rcos(p)+1, with the limits of integration set for r, t, and p. The correct spherical coordinate transformation is emphasized as x=rcos(t)sin(p), y=rsin(t)sin(p), and z=rcos(p).

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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 don't 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
 
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You have an error :
the spherical coordinate is:
x=rcos(t)sin(p)
y=rsin(t)sin(p)
z=rcos(p)
only!
 
Feynman said:
You have an error :
the spherical coordinate is:
x=rcos(t)sin(p)
y=rsin(t)sin(p)
z=rcos(p)
only!

my transformation deals with the area that the integral works on
urs deals with the integrand , both ways i don't know how to solve the integral
that we get
 

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