Solve the Integral in cylindrical coordinates(adsbygoogle = window.adsbygoogle || []).push({});

∫∫∫ 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).

I have these so far:

Using Cylindrical Coordinates,

∫∫∫B dV/√(x² + y² + (h-z)²)

= ∫(θ = 0 to 2π) ∫(z = 0 to R) ∫(r = -√(R² - z²) to √(R² - z²)) (r dr dz dθ) / √(r² + (h-z)²)

= 2π ∫(z = 0 to R) ∫(r = -√(R² - z²) to √(R² - z²)) r dr dz / √(r² + (h-z)²)

= 2 * 2π ∫(z = 0 to R) ∫(r = 0 to √(R² - z²)) r dr dz / √(r² + (h-z)²), by evenness

= 4π ∫(z = 0 to R) √(r² + (h - z)²) {for r = 0 to √(R² - z²)} dz

= 4π ∫(z = 0 to R) [√((R² - z²) + (h - z)²) - √(h - z)²] dz

= 4π ∫(z = 0 to R) [√((R² - z²) + (h - z)²) - (h - z)] dz, since h > R > r

= 4π ∫(z = 0 to R) [√(R² + h² - 2hz) - h + z] dz

= 4π [(-1/(3h)) (R² + h² - 2hz)^(3/2) - hz + z²/2] {for z = 0 to R}

= 4π {[(-1/(3h)) (R² + h² - 2hR)^(3/2) - hR + R²/2] - (-1/(3h)) (R² + h²)^(3/2)}

= 4π [(-1/(3h)) (h - R)³ - hR + R²/2 + (1/(3h)) (R² + h²)^(3/2)].

Now, is that solution correct? Is there a way so simplify things? And from there, how do I find the average?

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# Triple Integral and finding the average

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