The area of a disk enclosed by a circle of radius R is Pi*R2.
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*R3.
And the formula for the surface area of a sphere of radius R is 4*Pi*R2.
And, you can check that the latter is the derivative of the former with respect to R.
Coincidence, or is there a reason?
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 4-dimensional “volume” of the 4-dimensional ball is (1/2)*Pi2*R4, what is 3-dimensional 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. <https://www.math.hmc.edu/funfacts>.
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