Sketch the region $\{(x, y) : y = \sqrt{4 - x^2} \}$ and X-axis. Find the area of the region using integration.

$2\pi$ square units

The correct answer is Option (3) → $2\pi$ square units

Given region is $\{(x, y) : y = \sqrt{4 - x^2} \}$ and X-axis.

We have,

$y = \sqrt{4 - x^2} \Rightarrow y^2 = 4 - x^2 \Rightarrow x^2 + y^2 = 4 = (2)^2$

This is the equation of a circle with centre $(0, 0)$ and radius $2$.

$∴$ Area of shaded region $= \int_{-2}^{2} \sqrt{4 - x^2} \, dx = \int_{-2}^{2} \sqrt{2^2 - x^2} \, dx$

$= \left[ \frac{x}{2} \sqrt{2^2 - x^2} + \frac{2^2}{2} \sin^{-1} \frac{x}{2} \right]_{-2}^{2} \quad \left[ ∵\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} \right]$

$= \frac{2}{2} \sqrt{2^2 - 2^2} + 2 \sin^{-1} \frac{2}{2} - \left( \frac{-2}{2} \sqrt{2^2 - (-2)^2} + 2 \sin^{-1} \left( \frac{-2}{2} \right) \right)$

$= 0 + 2 \sin^{-1}(1) - (0 + 2 \sin^{-1}(-1))$

$= 2 \sin^{-1} \left( \sin \frac{\pi}{2} \right) - 2 \sin^{-1} \left( -\sin \frac{\pi}{2} \right)$

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

$= \pi - 2 \left( -\frac{\pi}{2} \right) = 2\pi \text{ sq. units}$