User:Lark046/Area

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

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

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

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.

${\begin{aligned}y&{}={\sqrt {r^{2}-x^{2}}}\\\mathrm {A} &{}\approx 4\sum y\bigtriangleup x\\&{}=4\int _{0}^{r}y\,dx\\&{}=4\int _{0}^{r}{\sqrt {r^{2}-x^{2}}}\,dx\end{aligned}}$

Let

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

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

Substituting:

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

Move the constants across the integral:

${\begin{aligned}\mathrm {A} &{}=4r^{2}\int _{0}^{\pi /2}sin^{2}\,\theta \,d\theta \end{aligned}}$

Let

${\begin{aligned}u=sin^{2}\,\theta \end{aligned}}$	${\begin{aligned}dv=d\theta \end{aligned}}$
${\begin{aligned}du&{}=2\,sin\,\theta \,cos\,\theta \,d\theta \\&{}=sin2\theta \,d\theta \\\end{aligned}}$	${\begin{aligned}v=\theta \end{aligned}}$

Substituting:

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

Let

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

Substituting:

${\begin{aligned}\mathrm {A} &{}=4r^{2}({\frac {\pi }{2}}-\int _{0}^{\pi /2}\theta \,sin\,2\theta \,d\theta )\\&{}=4r^{2}({\frac {\pi }{2}}-\int _{0}^{\pi }{\frac {\alpha }{2}}\,sin\,\alpha \,({\frac {1}{2}})d\alpha )\\\end{aligned}}$

Move the constants across the integral:

${\begin{aligned}\mathrm {A} &{}=4r^{2}({\frac {\pi }{2}}-{\frac {1}{4}}\int _{0}^{\pi }\alpha \,sin\,\alpha \,d\alpha )\\\end{aligned}}$

Let

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

Substituting:

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