Answer
$A=\pi\,r^2$
Work Step by Step
We can express the area of a circle by the double integral:
$$\iint_{R}dA$$
R represents the region inside a circle on the $xy$-plane centered at the origin with radius r.
We can rewrite this integral as an iterated integral by adding bounds and get:
$$\int_{-r}^{r}\bigg(\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}dy\bigg)dx$$
Cleaning up the inner integral, we get:
$$\int_{-r}^{r}\bigg(\bigg[y\bigg]_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\bigg)dx=\int_{-r}^{r}2\sqrt{r^2-x^2}\,dx$$
Because of symmetry, we can rewrite this integral as:
$$4\int_{0}^{r}\sqrt{r^2-x^2}\,dx$$
Using the substitution $x=r\sin\theta$ and changing the bounds accordingly, we can rewrite the integral as:
$$4\int_{0}^{\pi/2}\sqrt{r^2-r^2\sin^2\theta}\,r\cos\theta\,d\theta$$
Simplifying using the Pythagorean trigonometric identity, we get:
$$4r^2\int_{0}^{\pi/2}\cos^2\theta\,d\theta$$
Solving, we get:
$$4r^2\bigg[\frac{1}{2}+\frac{1}{4}\sin2x\bigg]_{0}^{\pi/2}\\=4r^2\bigg(\frac{\pi}{4}\bigg)=\pi r^2$$