Tirple Integral in a ball?

[pipc]

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

 

[pipc]

Is the following solution correct? Is there a way to simplify things?
And from then how do I take tha average?

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)].

 

[pipc]

I have to submit this paper, so can anyone please tell me if the answer is correct?

 

[renegade05]

Let me get this straight.

Your integral is the following:

\(\displaystyle \int\int\int_B\frac{1}{\sqrt{x^2+y^2+(h-z)^2}}}dxdydz\)

where B is the solid region bounded by \(\displaystyle x^2+y^2+z^2=R^2\)

And you want to convert his to cylindrical coordinates and solve the integral?