# Area of a circle

I've tried to find as a personnal exercise where the formula $A=\pi R^2$ comes from.

After drawing the problem, I've found that $A = 2\int\limits_{-R}^{R}\sqrt{R^2-t^2}dt$. How can I calculate this ?

• You mean $\pi R^2$. Feb 22, 2012 at 19:59
• You could try the substitution $t = R \cos(\theta)$. Do you know how to compute $\int \sin^2(\theta) d\theta$? Also, this should be tagged as calculus or geometry, not algebraic-geometry. (And, as mentioned above, you should correct the area formula.) Feb 22, 2012 at 20:00
• 1) The area of a circle is $0$. You are computing the area of a disc. 2) Your integral is off by a factor $2$ (it gives only the area of a half-disc). Feb 22, 2012 at 20:05
• Exactly @ÁlvaroLozano-Robledo, sorry for the little mistake... Michael, okay I'll try this way, thank you. Thomine : I forgot to write it but I had noticed it, thank you. Feb 22, 2012 at 20:06
• Depending on how one handles the basics, the proof may unfortunately be a little $\dots$ circular. The usual first-year calculus proof of the fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$ uses the area of the circle. From the limit calculation comes the derivative of $\sin x$, and hence the integral of $\cos^2 x$. Feb 22, 2012 at 21:44

This is a classic case of trigonometric substitution. Set $t=R\sin\theta$, $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$.
Then $R^2 - t^2 = R^2(1-\sin^2\theta) = R^2\cos^2\theta$, hence $\sqrt{R^2-t^2} = \sqrt{R^2\cos^2\theta} = |R\cos\theta| = R\cos\theta$, because $R\geq 0$ and $\cos\theta\geq 0$ for $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$.
Since $t=R\sin\theta$, $dt = R\cos\theta\,d\theta$. When $t=-R$, we have $\theta=-\frac{\pi}{2}$; when $t=R$, we have $\theta=\frac{\pi}{2}$. So we get $$\int_{-R}^R\sqrt{R^2-t^2}\,dt = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}R\cos^2\theta\,d\theta.$$
Now you can use the formula $$\int \cos^2\theta\,d\theta = \frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta+C$$ (which can be found by using integration by parts, then replacing $\sin^2\theta$ with $1-\cos^2\theta$, and "solving" for the integral of the cosine squared) to get the desired result.