Ok, I understand you're stuck in this assignment, but where exactly? What is your question to us?
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
Is the following solution looks OK?
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)].
And now, how do I find the average value?
No, you have the "r" and "z" limits of integration reversed (perhaps that's a typo). The sphere of radius R, with center (0, 0, 0), is given by . That is, for every r and , z varies from to . It is r that varies from 0 to R and, of course, varies from 0 to
Do you think the following solution is correct?
I got the coordinates a little different, is that still ok?
In cylindricals the ball is covered by [θ=0,2π], [z=−R,R], [r=0,√(R²−z²)]
∫ = ∫∫∫ rdθdrdz / √{r²+(z−h)²}, [θ=0,2π], [z=−R,R], [r=0,√(R²−z²)]
= 2π ∫∫ rdrdz / √{r²+(z−h)²}, [z=−R,R], [r=0,√(R²−z²)]
= 2π ∫ √{r²+(z−h)²} dz, [z=−R,R], [r=0,√(R²−z²)]
= 2π ∫ { √{R²−z²+(z−h)²} − √{(z−h)²} } dz, [z=−R,R]
= 2π ∫ { √{R²+h²−2zh} − (h−z) } dz, [z=−R,R] since h > z over range of z
= 2π { (−⅓h‾¹)(R²+h²−2zh)¹˙⁵ + ½(h−z)² }, [z=−R,R]
= 2π { (−⅓h‾¹)( (R²+h²−2Rh)¹˙⁵ − (R²+h²+2Rh)¹˙⁵ ) + ½(h−R)² − ½(h+R)² }
= 2π { (−⅓h‾¹)( (h−R)³ − (h+R)³ ) − 2hR }
= 2π { (−⅓h‾¹)(−6h²R−2R³) − 2hR } = 4πR³/(3h)