Area of a circle (Part 4)

Math majors are completely comfortable with the formula $A = \pi r^2$ 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 the previous three posts, I discussed various ways that calculus can be used to show that $A = \pi r^2$ . Still, most future high school teachers would like to know a justification for why $A = \pi r^2$ . After all, the definition of $\pi$ is

$\pi = \displaystyle \frac{ \hbox{Circumference} }{ \hbox{ Diameter}}$ , or $C = 2\pi r$

So there ought to be a reasonable explanation for why $\pi$ reappears in the formula for the area of a circle. Furthermore, this explanation should within the grasp of geometry students — so that the explanation should not explicitly use calculus. Even better, they’d prefer a hands-on classroom activity so that students could discover the formula for themselves.

The video below shows a completely geometric justification for why $A = \pi r^2$ that meets the above criteria. I have a couple of small quibbles with the narrated text — I’d prefer to say that the each rearrangement of pieces is approximately a parallelogram (as opposed to a rectangle), and that figures get closer and closer to a real parallelogram with area $A = \pi r^2$ .

In other words, I would avoid saying that we ultimately divide the circle into infinitely many wedges of infinitesimal width to get a perfect rectangle, as this promotes a misconception concerning the definition of a limit that they shouldn’t carry into a future calculus course.

However, the graphics are excellent in this video. In my mind, that more than counterbalances the preferred way that I would describe the process of taking a limit to students.

Pedagogically, I would recommend a hands-on activity along these lines. Let the students use a protractor to draw a 5- or 6-inch circle on a piece of paper. Then have them mark $18$ points on the circumference of the circle at every $20^o$ , and then draw the lines connecting these points and the center of the circle. Then have the students cut out these wedges and physically rearrange them as in the video. They should discover for themselves that the wedges approximately form a parallelogram, and they know how to find the area of a parallelogram.

After they do this activity, then I would show the above video to geometry students.

If anyone knows a video that (1) is as visually appealing as the one above and (2) correctly states the principle of limit for geometry students, please let me know.

Area of a circle (Part 3)

$A = \displaystyle \iint_R 1 \, dx \, dy$

This double integral may be computed by converting to polar coordinates. The distance from the origin varies from $r=0$ to $r=a$ , while the angle varies from $\theta = 0$ to $\theta = 2\pi$ . Using the conversion $dx \, dy = r \, dr \, d\theta$ , we see that

$A = \displaystyle \int_0^{2 \pi} \int_0^a r \, dr \, d \theta$

$A = \displaystyle \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_0^a \, d\theta$

$A = \displaystyle \int_0^{2\pi} \frac{a^2}{2} \, d\theta$

$A = \displaystyle 2 \pi \cdot \frac{a^2}{2}$

$A = \displaystyle \pi a^2$

We note that the above proof uses the fact that calculus with trigonometric functions must be done with radians and not degrees. In other words, we had to change the range of integration to $[0,2\pi]$ and not $[0^o, 360^o]$ .

Area of a circle (Part 2)

A circle centered at the origin with radius $r$ may be viewed as the region between $f(x) = -\sqrt{r^2 - x^2}$ and $g(x) = \sqrt{r^2 - x^2}$ . These two functions intersect at $x = r$ and $x = -r$ . Therefore, the area of the circle is the integral of the difference of the two functions:

$A = \displaystyle \int_{-r}^r \left[g(x) - f(x) \right] \, dx= \displaystyle \int_{-r}^r 2 \sqrt{r^2 - x^2} \, dx$

This may be evaluated by using the trigonometric substitution $x = r \sin \theta$ and changing the range of integration to $\theta = -\pi/2$ to $\theta = \pi/2$ . Since $dx = r \cos \theta \, d\theta$ , we find

$A = \displaystyle \int_{-\pi/2}^{\pi/2} 2 \sqrt{r^2 - r^2 \sin^2 \theta} \, r \cos \theta d\theta$

$A = \displaystyle \int_{-\pi/2}^{\pi/2} 2 r^2 \cos^2 \theta d\theta$

$A = \displaystyle r^2 \int_{-\pi/2}^{\pi/2} (1 + \cos 2\theta) d\theta$

$A = \displaystyle r^2 \left[ \theta + \frac{1}{2} \sin 2\theta \right]_{-\pi/2}^{\pi/2}$

$A = \displaystyle r^2 \left[ \left( \displaystyle \frac{\pi}{2} + \frac{1}{2} \sin \pi \right) - \left( - \frac{\pi}{2} + \frac{1}{2} \sin (-\pi) \right) \right]$

$A = \pi r^2$

Area of a circle (Part 1)

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 $A = \pi r^2$ .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

$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$

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

1. If $f(x) = c$ , a constant, then $\displaystyle \frac{d}{dx} (c) = 0$ .

2. If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$ .

3. If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$ .

4. If $f(x) = x^n$ , where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$ . This may be proved by at least two different techniques:

The binomial expansion $(x+h)^n = x^n + n x^{n-1} h + \displaystyle {n \choose 2} x^{n-2} h^2 + \dots + h^n$
The Product Rule (derived later) and mathematical induction

5. If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$ . 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 $A(r) = \pi r^2$ . Notice I’ve changed the variable from $x$ to $r$ , but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.) What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$ . Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$ . Does this remind you of anything? (Students answer: the volume of a sphere.) What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$ . 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 $A(r) = \pi r^2$ , then

$A'(r) = \displaystyle \lim_{h \to 0} \frac{ A(r+h) - A(r) }{h} = 2\pi r$

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

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

$A(r + h) - A(r) \approx 2 \pi r h$

and we can conclude that

$A'(r) = \displaystyle \lim_{h \to 0} \frac{ 2 \pi r h}{h} = 2\pi r$

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 $A'(r) = C(r)$ , as above. Therefore, by the Fundamental Theorem of Calculus,

$A(r) - A(0) = \displaystyle \int_0^r C(t) \, dt$

$A(r) - A(0) = \displaystyle \int_0^r 2\pi t \, dt$

$A(r) - A(0) = \displaystyle \left[ \pi t^2 \right]_0^r$

$A(r) - A(0) = \pi r^2$

Since the area of a circle with radius $0$ is $0$ , we conclude that $A(r) = \pi r^2$ .

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 $s$ , the area is $A(s) = s^2$ and the perimeter is $P(s) = 4s$ , which isn’t the derivative of $A(s)$ . The reason this didn’t work is because the side length $s$ of a square corresponds to the diameter of a circle, not the radius of a circle.

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

$A(x) = s^2 = (2x)^2 = 4x^2$

and

$P(x) = 4s = 4(2x) = 8x$

Written in terms of the half-sidelength $x$ , we see that $A'(x) = P(x)$ .

Collaborative Mathematics: Challenge 07

My colleague Jason Ermer has posted his 7th challenge video, shown below. It’s both an experiment and an exercise in probability.

Video responses can be posted to his website, http://www.collaborativemathematics.org. In the words of his website, this is a unique forum for connecting a worldwide community of mathematical problem-solvers, and I think these unorthodox but simply stated problems are a fun way for engaging students with the mathematical curriculum.

Collaborative Mathematics: Challenge 05

My colleague Jason Ermer is back from summer hiatus and has posted his fifth challenge video, shown below.

Engaging students: Equation of a circle

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Alyssa Dalling. Her topic, from Precalculus: the equation of a circle.

A. How could you as a teacher create an activity or project that involves your topic?

A fun way to engage students and also introduce the standard form of an equation of a circle is the following:

Start by separating the class into groups of 2 or 3
Pass each group a specific amount of flashcards. (Each group will have the same flashcards)
Each flashcard has a picture of a graphed circle and the equation of that circle in standard form
The students will work together to figure out how the pictures of the circle relate to the equation

This will help students understand how different aspects of a circle relate to its standard form equation. The following is an example of a flashcard that could be passed out.

Source: http://www.mathwarehouse.com/geometry/circle/equation-of-a-circle.php

C. How has this topic appeared in high culture (art, classical music, theatre, etc.)?

Circles have been used through history in many different works of art. One such type is called a tessellation. The word Tessellate means to cover a plane with a pattern in such a way as to leave no region uncovered. So, a tessellation is created when a shape or shapes are repeated over and over again. The pictures above show just a few examples of how circles are used in different types of art. A good way to engage students would be to show them a few examples of tessellations using circles.

Source: http://mathforum.org/sum95/suzanne/whattess.html

E. How can technology be used to effectively engage students with this topic?

Khan Academy has a really fun resource for using equations to graph circles. At the beginning of class, the teacher could allow students to play around with this program. It allows students to see an equation of a circle in standard form then they would graph the circle. It gives hints as well as the answer when students are ready. The good thing about this is that even if a student goes straight to the answer, they are still trying to identify the connection between the equation of the circle and the answer Khan Academy shows.

http://www.khanacademy.org/math/trigonometry/conics_precalc/circles-tutorial-precalc/e/graphing_circles