The area of a disk enclosed by a circle of radius R is Pi*R^{2}.
The formula for the circumference of a circle of radius R is 2*Pi*R.
A simple calculus check reveals that
the latter is the derivative of the former with respect
to R.
Similarly, the volume of a ball enclosed by a sphere of radius R is (4/3)*Pi*R^{3}.
And the formula for the surface area of a sphere of radius R is 4*Pi*R^{2}.
And, you can check that the latter is the derivative
of the former with respect to R.
Coincidence, or is there a reason?
Presentation Suggestions:
Let your students tell you those geometry formulas if they
remember them.
The Math Behind the Fact:
Well, no, it is not a coincidence. For the ball,
a small change in radius produces a
change in volume of the ball which is equal to the volume
of a spherical shell of radius R and thickness (delta R).
The spherical shell's volume is thus approximately
(surface area of the sphere)*(delta R).
But the derivative is approximately the
change in ball volume divided by (delta R), which is
thus just (surface area of the sphere).
So, if I tell you the 4dimensional "volume" of the
4dimensional ball is (1/2)*Pi^{2}*R^{4},
what is 3dimensional volume of its boundary?
See also Volume of a Ball in N Dimensions.
How to Cite this Page:
Su, Francis E., et al. "Surface Area of a Sphere."
Math Fun Facts.
<http://www.math.hmc.edu/funfacts>.
