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December 26th, 2011, 09:33 AM  #1 
Newbie Joined: Dec 2011 Posts: 4 Thanks: 0  Triple Integral in cylindrical coordinates
Solve the Integral in cylindrical coordinates ??? dxdydz/(sqrt( x^2 + y^2 + (hz)^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 
December 26th, 2011, 10:14 PM  #2 
Newbie Joined: Dec 2011 Posts: 4 Thanks: 0  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/?(x² + y² + (hz)²) = ?(? = 0 to 2?) ?(z = 0 to R) ?(r = ?(R²  z²) to ?(R²  z²)) (r dr dz d?) / ?(r² + (hz)²) = 2? ?(z = 0 to R) ?(r = ?(R²  z²) to ?(R²  z²)) r dr dz / ?(r² + (hz)²) = 2 * 2? ?(z = 0 to R) ?(r = 0 to ?(R²  z²)) r dr dz / ?(r² + (hz)²), by evenness = 4? ?(z = 0 to R) ?(r² + (h  z)²) {for r = 0 to ?(R²  z²)} dz = 4? ?(z = 0 to R) [?((R²  z²) + (h  z)²)  ?(h  z)²] dz = 4? ?(z = 0 to R) [?((R²  z²) + (h  z)²)  (h  z)] dz, since h > R > r = 4? ?(z = 0 to R) [?(R² + h²  2hz)  h + z] dz = 4? [(1/(3h)) (R² + h²  2hz)^(3/2)  hz + z²/2] {for z = 0 to R} = 4? {[(1/(3h)) (R² + h²  2hR)^(3/2)  hR + R²/2]  (1/(3h)) (R² + h²)^(3/2)} = 4? [(1/(3h)) (h  R)³  hR + R²/2 + (1/(3h)) (R² + h²)^(3/2)]. 

Tags 
coordinates, cylindrical, integral, triple 
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