Volume of Solid Ring

A solid sphere is bored out such that the radial axis of the removed cylinder passes through the centre. The ring of remaining material stands 6 centimetres high. What is a volume of this ring?

One way to calculate the volume of the ring is to subtract the volume of the removed material from the volume of the original sphere. This bored out material is able to be regarded as a right cylinder with spherical end caps on the two flat surfaces.

One more way is to compute the volume of the excluding the end caps, partial sphere, and then subtract the volume of the right cylinder. We will choose this method.

The volume of the partial sphere is able to be calculated by using the disk method and integrating from the top edge of the ring to the bottom edge. Using the centre of the sphere like the origin x as the horizontal axis through the centre and y the vertical axis the equation is

The right cylinder has volume

Vc = 6πr²

= 6π(R²- 9)

Where r is the radius of the cylinder Note the requirement: R ≥ 3. So the volume of the ring is V = V_s - V_c.

Subtracting

Astoundingly this result is independent of the radius of the sphere. As long as the radius R ≥ 3 the result holds. Therefore another way to compute the volume of the ring is to compute the volume of a sphere with R = 3 representing the case of an infinitesimal bored out volume. This sphere, of course, has volume given by

V =4/3πR³= 36π