Math majors are completely comfortable with the formula for the area of a circle. However, they often tell me that they don’t remember a proof or justification for why this formula is true. And they certainly don’t remember a justification that would be appropriate for showing geometry students.

In this series of posts, I’ll discuss several ways that the area of a circle can be found using calculus. I’ll also discuss a straightforward classroom activity by which students can discover for themselves why .In the first few weeks after a calculus class, after students are introduced to the concept of limits, the derivative is introduced for the first time… often as the slope of a tangent line to the curve. Here it is: if $y = f(x)$, then

From this definition, the first few rules of differentiation are derived in approximately the following order:

1. If , a constant, then .

2. If and are both differentiable, then .

3. If is differentiable and is a constant, then .

4. If , where is a nonnegative integer, then . This may be proved by at least two different techniques:

- The binomial expansion
- The Product Rule (derived later) and mathematical induction

5. If is a polynomial, then . In other words, taking the derivative of a polynomial is easy.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let . Notice I’ve changed the variable from to , but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.) What’s the derivative? Remember, is just a constant. So . Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Example 2. Now let’s try . Does this remind you of anything? (Students answer: the volume of a sphere.) What’s the derivative? Again, is just a constant. So . Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (

Students: stunned silence.)This is what’s known on television as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (

Students groan, as they really want to know the answer immediately.)

In the spirit of a cliff-hanger, I offer the following thought bubble before presenting the answer.

By definition, if , then

The numerator may be viewed as the area of the ring between concentric circles with radii and . In other words, imagine starting with a solid red disk of radius and then removing a solid white disk of radius . The picture would look something like this:

Notice that the ring has a thickness of . If this ring were to be “unpeeled” and flattened, it would approximately resemble a rectangle. The height of the rectangle would be , while the length of the rectangle would be the circumference of the circle. So

and we can conclude that

By the same reasoning, the derivative of the volume of a sphere ought to be the surface area of the sphere.

Pedagogically, I find that the above discussion helps reinforce the definition of a derivative at a time when students are most willing to forget about the formal definition in favor of the various rules of differentiation.

In the above work, we *started* with the formula for the area of the circle and then confirmed that its derivative matched the expected result. However, the above logic can be used to *derive* the formula for the area of a circle from the formula $C(r) = 2\pi r$ for the circumference. We begin with the observation that , as above. Therefore, by the Fundamental Theorem of Calculus,

Since the area of a circle with radius is , we conclude that .

Pedagogically, I don’t particularly recommend this approach, as I think students would find this explanation more confusing than the first approach. However, I can see that this could be useful for reinforcing the statement of the Fundamental Theorem of Calculus.

By the way, the above reasoning works for a square or cube also, but with a little twist. For a square of side length , the area is and the perimeter is , which isn’t the derivative of . The reason this didn’t work is because the side length of a square corresponds to the *diameter* of a circle, not the *radius* of a circle.

But, if we let denote half the side length of a square, then the above logic works out since

and

Written in terms of the half-sidelength , we see that .

After teaching differentiation to a group of students I always point out the if you differentiate the formula for the area of a circle, you get the formula for the circumference. This always amazes them, I’ll point them here next time they ask why!