Area of a circle (Part 2)

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 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.green line

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

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 [-\pi/2,\pi/2] and not [-90^o, 90^o].

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