The total area enclosed by the circle $x^2 + y^2 = a^2$ is

1. By symmetry the area is $4$ times the first-quadrant region: $A = 4\int_0^a \sqrt{a^2 - x^2}\,dx$. 2. Using $\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$, evaluate from $0$ to $a$. 3. At $x = a$ the bracket is $\frac{a^2}{2}\sin^{-1}1 = \frac{a^2}{2}\cdot\frac{\pi}{2}$. 4. So $A = 4\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi a^2$. _Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.2_