User:Lark046/Area

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

Proof that a circle’s area is πr2

\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

\begin{align} x &{}= r \, cos \, \theta \\ dx &{}= -r \, sin \, \theta \, d \theta \\ x^2 &{}= r^2 cos^2 \, \theta \\ r^2 - x^2 &{}= r^2 - r^2 cos^2 \, \theta \\ &{}= r^2(1 - cos^2 \, \theta) \\ &{}= r^2 sin^2 \theta \\ \sqrt{r^2 - x^2} &{}= r \, sin \, \theta \end{align}

\begin{align} 0 &{}= r \, cos \, \theta, r \neq 0 \\ 0 &{}= cos \, \theta \\ \frac{\pi}{2} &{}= \theta \\ r &{}= r \, cos \, \theta, r \neq 0 \\ 1 &{}= cos \, \theta \\ 0 &{}= \theta \\ \end{align}

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

\begin{align} u = sin^2 \, \theta \end{align}

\begin{align} dv = d \theta \end{align}

\begin{align} du &{}= 2 \, sin \, \theta \, cos \, \theta \, d \theta \\ &{}= sin 2 \theta \, d \theta \\ \end{align}

\begin{align} v = \theta \end{align}

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}

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