A Special Sphere in n-Dimensional Space
It's really interesting to consider objects that have their surface area equal to their volume.
For a circle of radius r in 2D we have surface area (circumference)= 2*pi*r and volume (area)=pi*r^2.
Equating these two gives r=2. Which is the dimension of the space!
Doing the same for a sphere in 3D we get r=3. Which is the dimension of the space!
Is this always true? Yes.
Take a point in n-dimensional Euclidean space (x1,x2,x3,...,xn)
Then the surface of a sphere of radius r is the set of points with:
and the volume consists of all the points with:
x1^2+x2^2+x3^2+....+xn^2 less than or equal to r^2
The volume of the sphere is c(n)*(r^n) and its surface area is n*c(n)*r^(n-1)
So equating them gives r=n.
Note that the function c(n) cancels out so its value is not needed, but c(n)=(pi^(n/2))/gamma(1+n/2). Where gamma is the Euler Gamma Function.
So we can say in general..
"In n-dimensional Euclidean space a sphere with surface area=volume has radius n"
The radius of the sphere is equal to the dimension of the space! I think this is a neat result because it relates the dimension of the space to a certain class of spheres within it. It may also be related to the Holographic Principle because it equates surface area and volume.
Content written and posted by Ken Abbott firstname.lastname@example.org