[list]p = pi

D = the diameter of the sphere

d = the diameter of the hole

R = the radius of the sphere (.5D)

r = the radius of the hole (.5d)

H = the height of the hole (end to end)

h = the height of the hole (centre to end = .5H)[list]

After drilling, the remaining ring shape is equivalent to a barrel of unknown width (D), 6" high (H) and with the central cylinder portion removed.

The formula for a barrel with sides bent to the arc of a circle

[list]= pH(2DD + dd) / 12

= (1/12)pH(8RR + 4rr) [list]

cpotting,

kudos for finding all these neat [cap and barrel] formulas.

That's does all of the calculus work.

I found it useful to put everything in terms of R and h.

You can do this by noting that RR = rr + hh [Pythagorus]

V[barrel] = ph [8RR + 4rr] /12 = ph [8RR + 4RR - 4hh]/6 = 2ph [RR - hh/3]

V[cylinder] = height x area = Hprr = 2ph [RR - hh]

V[barrel] - V[cylinder] = 2ph [RR - hh/3 - RR + hh] = 2ph [2hh/3] = 4phhh/3.

Recalling that h=3,

V[barrel] - V[cylinder] = 36p [a constant].

Looking at your derivation, everything is correct.

If you add like terms in your expression for V,

you'll see that all the rr terms add up to zero.

Your derivative expression is correct, also,

except that you should include the constants (1/12)pH8, etc...

[if y(x) = Cxx, then dy/dx = 2Cx, not just 2x]

You'll see again that the r-dependent terms add up to zero,

and the derivative is zero.

OK?