KCC's Quizzes AQQ281 about volume of a perforated sphere

From the rightmost picture:  (h/2)^2 = R^2 - r^2,  we find h knowing r and R.

Unfair (since I just took the formula from my CRC Handbook):
Volume of the sphere less it two caps:    2* (1/6) pi  (h/2) ( 3R^2 + 3r^2  + (h/2) ^2 ) 
                                                               = 2 pi (1/6) (h/2)  ( 4R^2  - 2r^2)      - -  after substitution of (h/2)^2

less volume of the cylinder:                       2 pi r^2 h

giving:    Volume of copper left after the drilling:    (1/3) pi h (2R^2 - 7r^2)    - - if my algebra is right.

Here the reference on Internet: CRC Standard Mathematical Tables and Formulas, 33rd Edition

The equation is at 4.9.1.2  and the figure on the previous page (223 of the book, 237 of the pdf)

Because the premise of the question is that the ratio of cylinder radius, r, vs sphere radius, R, is not relevant to the answer, we can picture the following progression:

where in the limit r=0, and R = h/2, so with h = 1, volume of the rightmost sphere is

4*pi*(1/2)^3 / 3 = pi/6 ~= 0.524 m^3

Now the hard part is proving the premise...

Going on to prove the premise, calculate the volume of the ring left after removing the cylinder.

Working with half the sphere, the volume left after cutting out the cylinder is half the sphere volume minus half the cylinder and the cap above the cylinder:

The cross sectional circle radius as a function of y is:

Putting it all together, half of the wedding band left when the cylinder is removed is:

and happily, this reduces to: pi * h^3 / 12

Doubling this and setting h=1 gives pi/6, matching the original result in the first post. (No, I didn't manually work out the integral, Mathcad -> simplify gave the reduced result. I'm too old to spend that much time on it.)

Wow, the book you mention is really fantastic! It will be very useful on many of our coming quizzes! Concerning this present quiz, you come thus with (1/3) pi h (2R^2 - 7r^2) . But the volume is dependent on r and R that are unknown...

Brilliant JohnFromL3Harris , you get the answer and with a very convincing development, congratulations!

Yes, I added instead of subtracting at one place.  It is easier working with half a sphere:

V0  = (pi/6) (h/2) ( 3r^2 +3R^2 + (h/2)^2)      - - half the sphere less its top, from CRC
V1 = ((pi/6) (h/2) (6r^2                           )      - - the matching cylinder; the 6/6  is to help factoring the common terms.

V0-V1 = (1/6) pi  (h/2) ( -3r^2 +3 R^2  +(R^2 - r^2) )  - - Pythagore, (h/2)^2 = R^2 - r^2          
           = (1/6)  pi (h/2) 2 ( 2R^2 - r^2)
Times 2 (back to the full sphere)
         => (2/3) pi h (2R^2 - r^2)                                   - - answer

testing limit cases:  r=0, full sphere, ok; 
                               r=R, h = 0, ok.

...  (2/3) pi h (2R^2 - r^2)   is reduced to pi/6 when h =1  since R^2 = (1/2)^2 + r^2

Answer is V=PI * h [R^2 - r^2 -h^2/12] = (PI*[R^2 - r^2 -1/12])

This comes from V= V_sphere - V_cylinder - 2 V_cap.

V_cap is the volume of the spherical caps generated by the cylinder with height = h/2<R and base radius=r

V_cap= (PI/3)*(b^2)*(3R-b), where b=R-h/2

V_sphere and V_cylinder can be expressed in terms of R, r and h. After a bit of algebra you get to result. 

In the limit as r_cylinder -> 0, the volume of the copper becomes (4/3)pi*R_sphere^3 = (4/3)pi*(h/2)^3 = pi*h^3/6.

This is a valid solution and no other constraints are given. Therefore, this is the solution.