The area of circle The question is to prove that area of a circle with radius $r$ is $\pi r^2$ using integral. I tried to write $$A=\int\limits_{-r}^{r}2\sqrt{r^2-x^2}\ dx$$
but I don't know what to do next.
 A: Hint: try a trigonometric substitution.  In particular, try setting $x = r \sin \theta$.
Also, note the identity:
$$
\cos^2 \theta = \frac 12 (1 + \cos(2\theta))
$$
A: $$2\sqrt{r^2-x^2}=2r\sqrt{1-\left(\frac xr\right)^2}$$
Substitute
$$x=r\sin t\;,\;\;dx=r\cos t\,dt\implies 2r\sqrt{1-\left(\frac xr\right)^2}dx=2r^2\cos^2t\,dt\implies$$
$$\int\limits_{-r}^r2\sqrt{r^2-x^2}dx=2r^2\int\limits_{-\pi/2}^{\pi/2}\cos^2t\,dt=\left.r^2\left(t+\sin t\cos t\right)\right|_{-\pi/2}^{\pi/2}=\pi r^2$$
A: Another interesting solution, not yet mentioned by the other answers, is to use this substitution:
$$
x = r\cos\theta \;\therefore\; \frac{dx}{d\theta} = -r\sin\theta
\;\therefore\; dx = -r\sin\theta \; d\theta
$$
Substituting the integral limits, we have 
$$x = -r \implies -r = r\cos\theta \implies \cos\theta = -1 \implies \theta = \pi$$
$$x = r \implies r = r\cos\theta \implies \cos\theta = 1 \implies \theta = 0$$
With this we can do:
$$
A=-r \int\limits_{\pi}^{0}2\sqrt{r^2-(r\cos\theta)^2} \sin\theta \; d\theta = 
-2r \int\limits_{\pi}^{0}\sqrt{r^2(1 - \cos^2\theta)} \sin\theta \; d\theta = 
$$
$$
-2r \int\limits_{\pi}^{0}\sqrt{r^2\sin^2\theta} \sin\theta \; d\theta = 
-2r \int\limits_{\pi}^{0}r\sin^2\theta \; d\theta = 
-2r^2 \int\limits_{\pi}^{0}\sin^2\theta \; d\theta = 
$$
$$
-2r^2 \int\limits_{\pi}^{0}\frac{1}{2}(1 - \cos(2\theta)) \; d\theta = 
$$
Now we substitute $u = 2\theta \implies \frac{du}{d\theta} = 2 \implies d\theta = \frac{du}{2}$, and limits range from $2\pi$ to $0$:
$$
\frac{-r^2}{2} \int\limits_{2\pi}^{0}(1 - \cos u) \; du = 
\frac{-r^2}{2} (u - \sin u)\Big{|}_{2\pi}^{0} = 
$$
$$
\frac{-r^2}{2} [(0 - \sin 0) - (2\pi - \sin2\pi)] = 
\frac{-r^2}{2} [-2\pi] = \pi r^2
$$
