Unizor - Geometry3D - Spheres - Problems 2

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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