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December 26th, 2011, 10:33 AM   #1
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Triple Integral in cylindrical coordinates

Solve the Integral in cylindrical coordinates
??? dxdydz/(sqrt( x^2 + y^2 + (h-z)^2)
B

Where B is the Ball with a Radius R around (0,0,0), and the parameter h is greater than R.


And then infer the average on that ball B with radius R of the distance opposite to the point (0,0,h)

(the average of function f on some shape v is defined as
[ ??? f(x,y,z) dxdydz ] / vol V
v
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December 26th, 2011, 11:14 PM   #2
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Re: Triple Integral in cylindrical coordinates

I did some work, I still don't know how to find the average.

Is the following correct?
Using Cylindrical Coordinates,
???B dV/?(x + y + (h-z))
= ?(? = 0 to 2?) ?(z = 0 to R) ?(r = -?(R - z) to ?(R - z)) (r dr dz d?) / ?(r + (h-z))
= 2? ?(z = 0 to R) ?(r = -?(R - z) to ?(R - z)) r dr dz / ?(r + (h-z))
= 2 * 2? ?(z = 0 to R) ?(r = 0 to ?(R - z)) r dr dz / ?(r + (h-z)), by evenness
= 4? ?(z = 0 to R) ?(r + (h - z)) {for r = 0 to ?(R - z)} dz
= 4? ?(z = 0 to R) [?((R - z) + (h - z)) - ?(h - z)] dz
= 4? ?(z = 0 to R) [?((R - z) + (h - z)) - (h - z)] dz, since h > R > r
= 4? ?(z = 0 to R) [?(R + h - 2hz) - h + z] dz
= 4? [(-1/(3h)) (R + h - 2hz)^(3/2) - hz + z/2] {for z = 0 to R}
= 4? {[(-1/(3h)) (R + h - 2hR)^(3/2) - hR + R/2] - (-1/(3h)) (R + h)^(3/2)}
= 4? [(-1/(3h)) (h - R) - hR + R/2 + (1/(3h)) (R + h)^(3/2)].
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