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