## Monday, November 2, 2015

### Unizor - Geometry3D - Sphere - Volume and Surface Area

Unizor - Creative Minds through Art of Mathematics - Math4Teens

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

Volume and Surface Area of a Sphere

Volume

Let's start with a volume of a sphere of a radius R and a center at point O using already familiar process of approximation of this volume as a sum of volumes of other geometric objects with known formulas for volume. We will use cylinders of the same but small altitude and different radiuses stacked on each other as an approximation of a sphere.

Imagine a plane that cuts our sphere of a radius R in two equal halves by going through its center. We will calculate the volume of a half a sphere and then double it.

From a center of a base O draw a perpendicular to this plane "upward" to intersection with our half a sphere at point P on its "top". The length of this perpendicular OP is, of course, R, since this is a radius of an original sphere.

Divide this perpendicular into N equal parts by points A1, A2,,, AN-1 (for uniformity, we can designate point P at the end of this perpendicular as AN) and draw planes parallel to a base through each such division point. The intersection of each plane and a sphere is a circle with a center at a corresponding division point Ak.

The next step is to make a cylinder within each layer preserving it's circular top base and replacing the side surface with a cylindrical surface by dropping perpendiculars from each point on the upper base towards its bottom base.

By Pythagorean theorem, for any point X on a circle with center Ak
AkX² = OX² − OAk²
Point X is on a sphere, therefore OX=R
Point Ak is kth point of division of a radius OP into N equal parts, therefore OAk=R·k/N.
Hence, the square of a radius of a base of the kth cylinder equals to
AkX² = R²−(R·k/N)²
The above is a radius of the kth cylinder.
Its altitude is, obviously, R/N.
Therefore, the volume of the kth cylinder is
Vk = π[R²−(R·k/N)²]·R/N =
= πR³/N − πR³·k²/N³
Now we have to summarize the volumes of all N cylinders and find the limit of this sum as N→∞. Doubling this (since we were dealing with half a sphere) will give us the volume of a sphere.
The result, the volume of the whole sphere of radius R will then be
Vsphere = 4πR³/3

Surface Area

Let's inscribe a convex polyhedron into a sphere by choosing a sufficiently large number of points on its surface (they will be vertices of a polyhedron) and connecting each point with its closest neighbors (these connections will be edges of the polyhedron and each triangle formed by these edges will be its face).
Connecting all the vertices with a center of a sphere, we divide a polyhedron into pyramids with different faces and, generally speaking, different altitudes.

As we add new points on a sphere, evenly distributing them in the empty spots on a sphere's surface, we will increase the number of faces of these polyhedrons, increasing the number of pyramids these polyhedrons are divided. What's interesting is that altitudes of all the pyramids will be closer and closer to a radius of a sphere and the surface area of a polyhedron becomes closer and closer to a surface area of sphere.

In the limit, as all faces are getting smaller and smaller, and the distance between a center of a sphere and all these faces becomes closer and closer to a radius of a sphere, we can say that the volume of a polyhedron (that approximates the volume of a sphere) approximately equals to one third of its surface area (that approximates the surface area of a sphere) multiplied by the radius of a sphere (that approximates the altitudes of all pyramids). The approximation is better as the largest face area of a polyhedron tends to zero.

The above logic is an intuitive base for the following statement about the relationship between the volume of a sphere Vsphere of a radius R and its surface area Ssphere:
Vsphere = Ssphere·R/3

Knowing the formula for the volume of a sphere, we can derive from this the formula for the surface area:
4πR³/3 = Ssphere·R/3
form which immediately follows:
Ssphere = 4πR²