Area of Circle

The area of a circle can be calculated by integration and use of trigonometric identities.

'''NOTE: There is a fault somewhere in the integration. Without this fault, the correct fulmula $$A=r^2\pi$$ would be produced.'''

From the circle equation $$x^2+y^2=r^2$$, we can determine that the area $$A$$ of the circle ois given by $$A/4=\int_0^r \sqrt{r^2-x^2}dx$$.

$$A/4=r\int_0^r \sqrt{1-\frac{x^2}{r^2}}dx$$

$$u=\frac{x}{r}, du=r^{-1}dx$$

$$A/4=\int_0^1 \sqrt{1-u^2}du$$

$$\sin v =x, \cos v dv = du$$

$$A/4=\int_0^{\sin(1)} \sqrt{1-\sin^2v}\cos v dv = \int_0^{\sin 1}\cos^2 v dv$$

$$\cos^2 v = \frac{1}{2}\cos(2v)+\frac{1}{2}$$

$$A/4=\frac{1}{2}\int_0^{\sin 1}\cos(2v)+1dv=\frac{1}{4}\sin(2v)+\frac{1}{2}v+C |_0^{\sin 1}$$

$$\sin(2v)=2\cos(v)\sin(v)$$

$$A/4=\frac{1}{2}\cos(v)\sin(v)+ \frac{1}{2}v+C |_0^{\sin 1}=\frac{1}{2}\sqrt{1-\sin^2v}\sin{v}+\frac{1}{2}v+C |_0^{\sin 1}$$

$$A/4=\frac{1}{2}\sqrt{1-u^2}u+\frac{1}{2}\arcsin u |_0^1$$

$$A/4=\frac{x}{2r}\sqrt{1-\frac{x^2}{r^2}}+\frac{r}{2}\arcsin(\frac{x}{r}) |_0^r$$

$$A/4=\frac{1}{2}\cdot 0 + \frac{r}{2}\arcsin(1)- 0\cdot 1 - \frac{r}{2}\arcsin(0)=\frac{r}{2}(\arcsin(1)-\arcsin(0)$$

$$A/4=\frac{r}{4}\pi$$

$$A=r\pi$$