Area of a circle by integration

Question

So I've been experimenting with new maths and I wanted to know if what I've done is valid. To start the equation for a circle is $x^2+y^2=r^2$ so $y^2=r^2-x^2$ and $y=\pm\sqrt{r^2-x^2}$ therefore: $$Area=2\int_{-r}^r\sqrt{r^2-x^2}dx$$ I'll spare you the integration $$Area = 2\biggl[\frac{\arcsin(x)+x\sqrt{r^2-x^2}}{2}\biggr]_{-r}^r$$ $$Area = \arcsin(r)-\arcsin(-r)+r\sqrt{r^2-r^2}-\biggl(-r\sqrt{r^2-(-r^2)}\biggr)$$ $$Area = 2\arcsin(r)+0+0$$ $$Area = 2\arcsin(r)$$ So far so good (using radians obviously) but it doesn't work for anything bigger than r=1 as arcsin is not defined. Here is my best go but it doesn't work I don't think. $$e^{i\theta}=\cos\theta+i\sin\theta$$ $$e^{-i\theta}=\cos\theta-i\sin\theta$$ $$e^{i\theta}-e^{-i\theta}=2i\sin\theta$$ $$\frac{e^{i\theta}-e^{-i\theta}}{2i}=\sin\theta$$ $$\frac{ie^{-i\theta}-ie^{i\theta}}{2}=\sin\theta$$ $$\frac{ie^{-i\theta}-ie^{i\theta}}{2}=r$$ $$\frac{1}{ie^{i\theta}}-ie^{i\theta}=2r$$ $$\frac{1-ie^{2i\theta}}{ie^{i\theta}}=2r$$ $$1-ie^{2i\theta}=2rie^{i\theta}$$ $$ie^{2i\theta}+2rie^{i\theta}-1=0$$ $$(ie^{i\theta}+2r)^2-4r^2-1=0$$ $$(ie^{i\theta}+2r)^2=4r^2+1$$ $$ie^{i\theta}+2r=\pm\sqrt{4r^2+1}$$ $$ie^{i\theta}=-2r\pm\sqrt{4r^2+1}$$ $$e^{i\theta}=(2r\mp\sqrt{4r^2+1})i$$ $$e^{i\theta}=(2r\mp\sqrt{4r^2+1})e^{\frac{i\pi}{2}}$$ $$i\theta=\ln(2r\mp\sqrt{4r^2+1})+\frac{i\pi}{2}$$ $$\theta=\frac{\pi}{2}-\ln(2r\mp\sqrt{4r^2+1})i$$ Therefore as $\arcsin(r)=\theta$ $$\arcsin(r)=\frac{\pi}{2}-\ln(2r\mp\sqrt{4r^2+1})i$$ $$Area = \pi-2\ln(2r\mp\sqrt{4r^2+1})i$$ But this doesn't work in the slightest ... Thank you for reading and I hope someone out there can fix this. It's just a hobby thing I do with lots of research so I did not come up with some of the initial ideas myself but I tried my hardest to try and do as much as I could.

Answer 1

Your integration is wrong. If you want to do the indefinite integral, $$ 2\int \sqrt{r^2-x^2} \, dx, $$ you set $x=r\sin{\theta}$, so $dx=r\cos{\theta} \, d\theta $, and the integral becomes $$ 2r^2\int \cos^2{\theta} \, d\theta = r^2\int (1+\cos{2\theta}) \, d\theta = r^2\theta+\frac{r^2}{2}\sin{2\theta} = r^2\arcsin{\left(\frac{x}{r}\right)} + x\sqrt{r^2-x^2} $$ This is continuous for $-r \leq x \leq r$, so we can use it to evaluate the definite integral, $$ \int \sqrt{r^2-x^2} \, dx = \left. r^2\arcsin{\left(\frac{x}{r}\right)} + x\sqrt{r^2-x^2} \right|_{-r}^r \\ = r^2\arcsin{1} + 0 - r^2\arcsin{-1} + 0 = 2r^2\arcsin{1} $$ since $\arcsin{y}$ is odd. And of course $\arcsin{1}=\pi/2$, so you get $\pi r^2$.

Answer 2

$$ Area=2\int_{-r}^r\sqrt{r^2-x^2}dx$$

$$= 2r^2\int_{-\pi/2}^{\pi/2}\cos ^2 (\theta ) d\theta $$

$$=r^2\int_{-\pi/2}^{\pi/2}\ ( 1+ \cos 2\theta ) d\theta $$

$$=r^2(\theta + 1/2 \sin 2\theta)|_{-\pi/2}^{\pi /2} = \pi r^2$$