User:Lark046/Area

March 27, 2007 (originally conceived April 25, 2001):

Proof that a circle’s area is $$ \pi r^2$$

$$\begin{align} r^2 &{}= x^2 + y^2 \\ y^2 &{}= r^2 - x^2 \\ y &{}= \pm \sqrt{r^2 - x^2} \end{align}$$

Looking at only the first quadrant, we can constrain $$y > 0$$. To get the total area, take the area found in the first quadrant and multiply by 4.

$$\begin{align} y &{}= \sqrt{r^2 - x^2} \\ \mathrm A &{}\approx 4 \sum y \bigtriangleup x \\ &{}= 4 \int_0^r y \, dx \\ &{}= 4 \int_0^r \sqrt{r^2 - x^2} \, dx \end{align}$$

Let

Substituting:

$$\begin{align} \mathrm A &{}= 4 \int_0^r \sqrt{r^2 - x^2} \, dx \\ &{}= 4 \int_{\pi/2}^0 (r \, sin \, \theta) (-r \, sin \, \theta \, d \theta) \\ &{}= 4 \int_{\pi/2}^0 -r^2 \, sin^2 \, \theta \, d \theta \\ \end{align}$$

Move the constants across the integral:

$$\begin{align} \mathrm A &{}= 4r^2 \int_0^{\pi/2} sin^2 \, \theta \, d \theta \end{align}$$

Let

Substituting:

$$\begin{align} \mathrm A &{}= 4r^2 \int_0^{\pi/2} sin^2 \, \theta d \theta \\ &{}= 4r^2 (\theta \, sin^2 \, \theta |_0^{\pi/2} - \int_0^{\pi/2} \theta \, sin \, 2 \theta \, d \theta) \\ &{}= 4r^2 ((\frac{\pi}{2})(1^2) - (0)(0^2) - \int_0^{\pi/2} \theta \, sin \, 2 \theta \, d \theta) \\ &{}= 4r^2 (\frac{\pi}{2} - \int_0^{\pi/2} \theta \, sin \, 2 \theta \, d \theta) \\ \end{align}$$

Let

$$\begin{align} \alpha &{}= 2 \theta \\ d \alpha &{}= 2 \, d \theta \\ \theta &{}= \frac{\alpha}{2} \\ 0 &{}= \frac{\alpha}{2}, 0 = \alpha \\ \frac{\pi}{2} &{}= \frac{\alpha}{2}, \pi = \alpha \\ \end{align}$$

Substituting:

$$\begin{align} \mathrm A &{}= 4r^2 (\frac{\pi}{2} - \int_0^{\pi/2} \theta \, sin \, 2 \theta \, d \theta) \\ &{}= 4r^2 (\frac{\pi}{2} - \int_0^{\pi} \frac{\alpha}{2} \, sin \, \alpha \, (\frac{1}{2})d \alpha) \\ \end{align}$$

Move the constants across the integral:

$$\begin{align} \mathrm A &{}= 4r^2 (\frac{\pi}{2} - \frac{1}{4} \int_0^{\pi}\alpha \, sin \, \alpha \, d \alpha) \\ \end{align}$$

Let

$$\begin{align} u &{}= \alpha, \, dv = sin \, \alpha \, d \alpha \\ du &{}= d \alpha, v = -cos \, \alpha \\ \end{align}$$

Substituting:

$$\begin{align} \mathrm A &{}= 4r^2 (\frac{\pi}{2} - \frac{1}{4} \int_0^{\pi} \alpha \, sin \, \alpha \, d \alpha) \\ &{}= 4r^2 (\frac{\pi}{2} - \frac{1}{4} [\alpha \, cos \, \alpha |_0^{\pi} - \int_{\pi}^0 -cos \, \alpha \, d \alpha]) \\ &{}= 4r^2 (\frac{\pi}{2} - \frac{1}{4} [(\pi)(1) - (0)(1) - \int_0^{\pi} cos \, \alpha \, d \alpha]) \\ &{}= 4r^2 (\frac{\pi}{2} - \frac{1}{4} [\pi - sin \, \alpha |_0^{\pi}]) \\ &{}= 4r^2 (\frac{\pi}{2} - \frac{1}{4} [\pi - (0 - 0)]) \\ &{}= 4r^2 (\frac{\pi}{2} - \frac{\pi}{4}) \\ &{}= 4r^2 (\frac{\pi}{4}) \\ \mathrm A &{}= \pi r^2 \end{align}$$