## Wednesday, November 25, 2015

### Unizor - Geometry3D - Spheres - Problems 2

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spheres - Problems 2

Just as a reminder, here are a few formulas related to spheres. They are all derived in the previous lectures and are helpful in solving these problems.

Volume of a sphere: 4πR³/3

Surface area of a sphere: 4πR²

Volume of a cap: πH²(3R−H)/3

Surface area of a dome: 2πRH

Volume of a sector: 2πR²H/3

Problem

Given a sphere of some unknown radius.

A cylindrical hole is drilled through it such that the axis of a cylinder goes through a center of a sphere. The height of a hole from edge to edge is h.

What is the volume of remaining part of a sphere?

Answer

V = πh³/6

Solution

The drilled out part of a sphere is a cylinder with two caps on both ends.

To calculate their volumes we need the radius of a sphere R, the radius of a cylinder r, the height of each cap H and the heights of a cylinder h with only the height of a cylinder h given.

Challenging, is not it?

From Pythagorean Theorem:

R² = r² + (h/2)²

The heights of a cylinder and caps is related to the radius of a circle:

R = H+h/2

So, we have only two equations with three unknown - R, r and H.

Let's express r² in terms of known h and unknown H:

r² = (H+h/2)²−(h/2)² = H²+Hh

Let's proceed with calculations of a volume of the remaining after drilling part of a sphere, hoping that the resulting formula would be a function of only cylinder's height h.

The volume of a sphere is

Vsphere = 4πR³/3

The volume of a cylinder is

Vcyl = πr²h

The volume of each cap is

Vcap = πH²(3R−H)/3

The volume of a remaining after drilling part of a sphere is

V = Vsphere − Vcyl − 2Vcap =

=4πR³/3−πr²h−2πH²(3R−H)/3

Let's substitute in this expression

R = H+h/2 and

r² = H²+Hh

The result will be:

V = 4π(H+h/2)³/3 −

− π(H²+Hh)h −

− 2πH²[3(H+h/2)−H]/3 =

= π{4H³+6H²h+3Hh²+h³/2 −

− 3H²h−3Hh² −

− 4H³−3H²h}/3 =

= πh³/6

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