My Math Forum Triple Integral in cylindrical coordinates

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 December 26th, 2011, 10:33 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 + (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
 December 26th, 2011, 11:14 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 + (h-z)²) = ?(? = 0 to 2?) ?(z = 0 to R) ?(r = -?(R² - z²) to ?(R² - z²)) (r dr dz d?) / ?(r² + (h-z)²) = 2? ?(z = 0 to R) ?(r = -?(R² - z²) to ?(R² - z²)) r dr dz / ?(r² + (h-z)²) = 2 * 2? ?(z = 0 to R) ?(r = 0 to ?(R² - z²)) r dr dz / ?(r² + (h-z)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)].

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