A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 9...?

Add. Dets.: You don't have to subtract the caps if you don't include them in the first place.

Call

the radius of the sphere, r

the radius of the drill, a

Put the origin of coordinates at the center of the sphere, and the z-axis along the axis of the drill.

After drilling, the remaining solid goes from z = -c to c, where

c^2 = r^2 - a^2

Its cross-section at a plane of constant z is an annulus with:

inner radius a, and

outer radius b = √(r^2 - z^2)

so its cross-sectional area is

A(z) = π(b^2 - a^2) = π(r^2 - a^2 - z^2) = π(c^2 - z^2)

The volume of the solid is then

V = ∫[z=-c,c] A(z)dz

= 2∫[z=0,c] A(z)dz

= 2π ∫[z=0,c] (c^2 - z^2)dz

= 2π (c^2 z - z^3 /3) [z=0,c]

= 2π [c^2 c - c^3 /3]

= 2π (1 - 1/3)c^3

= (4π/3)c^3

Sanity checks:

when a=0, c = r, and this is just the volume of the sphere ... √

when a=r, c = 0, V=0 ... √

So for a = 4, r = 9,

c^2 = 81 - 16 = 65

V = (4π/3)c^3 = (4π/3)65^(3/2) = 260π√65 /3 = 2195.1219...

EDIT:

sda, Kdr: Yes, sda, Craig has the wrong formula for the sphere, but you both have the wrong formula for this figure -- If the drill radius were 8 or even 7.35, your method would yield a negative volume!

Kdr: Your cylinder needs a radius other than r, which you've chosen as the radius of the sphere. Call it a, and you have

V = 4π/3 r^3 - π a^2 h

and when you take h = 2r, this becomes

V = 2π r((2/3)r^2 - a^2)

So when a satisfies

2/3 < (a/r)^2 < 1,

you'll get V < 0, when it should be > 0

Hint: a cylinder of height 2*9=18 units cannot lie entirely inside a sphere of diameter=18 units!

V = volume of sphere - volume of cylinder

V = [4/3 * pi * r^3] - [pi * r^2 * h ]

V = (4/3 * 22/7 * 9^3) - ( 22/7 * 4^2 * 18 )

you have to subtract the inner volume from the outer volume.

(4/3)pi*(9)^2- (4/3)pi*(4^2)

(4/3)pi[81-16]

(4/3)pi[80]

answer: 320pi/3 cubic units.

4/3π9^3; π4^2(18)

972π-288π

684π=v

Vsphere - (Vcylinder + 2Vcaps)