Find the area enclosed by the circle $x^2 + y^2 = a^2$.

$\pi a^2$

The correct answer is Option (1) → 1

From given figure, the whole area enclosed by the given circle

$= 4 (\text{area of the region AOBA bounded by the curve, } x\text{-axis and the ordinates } x = 0 \text{ and } x = a)$ [as the circle is symmetrical about both $x$-axis and $y$-axis]

$= 4 \int\limits_{0}^{a} y \, dx \quad \text{(taking vertical strips)}$

$= 4 \int\limits_{0}^{a} \sqrt{a^2 - x^2} \, dx$

Since $x^2 + y^2 = a^2$ gives $y = \pm\sqrt{a^2 - x^2}$.

As the region AOBA lies in the first quadrant, $y$ is taken as positive. Integrating, we get the whole area enclosed by the given circle:

$= 4 \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1} \frac{x}{a} \right]_{0}^{a}$

$= 4 \left[ \left( \frac{a}{2} \times 0 + \frac{a^2}{2}\sin^{-1} 1 \right) - 0 \right] = 4 \left( \frac{a^2}{2} \right) \left( \frac{\pi}{2} \right) = \pi a^2$