Starting from:

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

The area of the circle is therefore:

$$\begin{aligned} \int_{-r}^{r} 2 \sqrt{r^2 - x^2} \, dx \\ \end{aligned}$$

Let $\displaystyle \, x = r\sin \theta$, so $dx = r \cdot \cos{\theta} \, d\theta$:

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

Let $\displaystyle \, u = 2 \theta$, so $du = 2 \, d\theta$:

$$\begin{aligned} r^2 \int_{-\pi/2}^{\pi/2} (1 + \cos 2{\theta}) \, d\theta &= \frac{r^2}{2} \int_{-\pi}^{\pi} (1 + \cos{u}) \, du \\ &= \frac{r^2}{2} (u + \sin u) \, \biggr|_{-\pi}^{\pi} \\ &= \pi \, r^2 \\ \end{aligned}$$

$\Box$

# posted by rot13(Unafba Pune) @ 4:47 PM