### Brain Teasers

# Drilling Through A Sphere

I just drilled straight through the center of a solid sphere, resulting in a 6cm long cylindrical hole. What is the volume remaining in the sphere?

Here are some formulas you may find useful:

Volume of sphere = (4/3)Pi(r^3)

Volume of cap = (1/6)Pi(3R^2+h^2)h

Volume of cylinder = Pi(R^2)H

r = radius of sphere

h = height of cap

R = radius of hole (the cap and the cylinder share this radius)

H = height of cylinder (6cm)

The portion of the sphere that was removed can be thought of as a cylinder with a cap at both ends. A cap is flat on one side (touching the cylinder) and spherical on the other side (the original surface of the sphere).

Here are some formulas you may find useful:

Volume of sphere = (4/3)Pi(r^3)

Volume of cap = (1/6)Pi(3R^2+h^2)h

Volume of cylinder = Pi(R^2)H

r = radius of sphere

h = height of cap

R = radius of hole (the cap and the cylinder share this radius)

H = height of cylinder (6cm)

The portion of the sphere that was removed can be thought of as a cylinder with a cap at both ends. A cap is flat on one side (touching the cylinder) and spherical on the other side (the original surface of the sphere).

### Hint

Hint for a hard solution:The sphere's diameter (2r) must be the same as the length of the cylinder (H) plus the height of the two caps (2h), so 2r=H+2h.

A right-angled triangle is formed by the radius of the cylinder (R) and half the height of the cylinder (H/2), with the radius of the sphere (r) forming the hypotenuse. Using Pythagoras, R^2+(H/2)^2=r^2.

Hint for an easy solution:

Assume there is a real answer. Therefore, the answer does not depend on the radius of the hole (R). Use whatever value is easiest.

### Answer

The volume is exactly 36Pi cm^3 (about 113.1 cm^3).Hard solution:

The sphere's diameter (2r) must be the same as the length of the cylinder (H) plus the height of the two caps (2h), so 2r=H+2h.

Thus, 2h=2r-H and H=6, so 2h=2r-6 or simply h=r-3.

A right-angled triangle is formed by the radius of the cylinder (R) and half the height of the cylinder (H/2), with the radius of the sphere (r) forming the hypotenuse. Using Pythagoras: R^2+(H/2)^2=r^2.

Thus, R^2=r^2-(H/2)^2 and H=6, so R^2=r^2-9.

Volume of cap = (1/6)Pi(3R^2+h^2)h

(replace R^2 and h with values computed above)

= (1/6)Pi(3(r^2-9)+(r-3)^2)(r-3)

= (1/6)Pi(3r^2-27+r^2-6r+9)(r-3)

= (1/6)Pi(4r^2-6r-18)(r-3)

= (1/6)Pi(4r^3-6r^2-18r-12r^2+18r+54)

= (1/6)Pi(4r^3-18r^2+54)

= (2/3)Pi(r^3)-3Pi(r^2)+9Pi

Volume of cylinder = Pi(R^2)H

(replace R^2 with value computed above, and H with given value of 6)

= Pi(r^2-9)6

= 6Pi(r^2)-54Pi

Volume of sphere remaining = sphere - 2 caps - cylinder

= [ (4/3)Pi(r^3) ] - [ 2 ((2/3)Pi(r^3)-3Pi(r^2)+9Pi) ] - [ 6Pi(r^2)-54Pi ]

= [ (4/3)Pi(r^3) ] - [ (4/3)Pi(r^3)-6Pi(r^2)+18Pi ] - [ 6Pi(r^2) - 54Pi ]

= (4/3)Pi(r^3) - (4/3)Pi(r^3) + 6Pi(r^2) - 18Pi - 6Pi(r^2) + 54Pi

= 36Pi

Easy solution:

Assume the hole is infinitely small (i.e. a radius of 0). Then, the volume of the cylinder (and the volume of the caps) is 0. The height of the cap is 0, so the radius of the sphere (r) is 3cm (half of the length of the cylindrical hole). Thus:

Volume of sphere remaining = sphere - 2 caps - cylinder

= (4/3)Pi(r^3) - 0 - 0

= (4/3)Pi(3^3)

= 36Pi

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