# Thread: Triple integral in a ball?

1. ## Triple integral in a ball?

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

2. ## re: Triple integral in a ball?

Ok, I understand you're stuck in this assignment, but where exactly? What is your question to us?

3. ## re: Triple integral in a ball?

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?

4. ## Re: Triple integral in a ball?

I have to specifically solve with Cylindrical Coordinates, but I am always confused with them, did I do an OK job with them in my solution? And is my final term is correct?

5. ## Re: Triple integral in a ball?

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 $x^2+ y^2+ z^2= r^2+ z^2= R^2$. That is, for every r and $\theta$, z varies from $z= -\sqrt{R^2- r^2}$ to $z= \sqrt{R^2- r^2}$. It is r that varies from 0 to R and, of course, $\theta$ varies from 0 to $2\pi$

6. ## Re: Triple integral in a ball?

I see, thanks, but, hmm, the final answer is still correct, no?

Or should I do all the calculations over again?

7. ## Re: Triple integral in a ball?

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)