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 4-dimensional “volume” of the 4-dimensional ball is (1/2)*Pi^{2}*R^{4}, 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>.

**Fun Fact suggested by: **

Francis Su