Area of the region in the first quadrant enclosed by the X-axis, the line $y = x$ and the circle $x^2 + y^2 = 32$ is

$4\pi$ sq units

The correct answer is Option (2) → $4\pi$ sq units

We have, area enclosed by X-axis i.e., $y = 0$, $y = x$ and the circle $x^2 + y^2 = 32$ in first quadrant.

Since, $x^2 + (x)^2 = 32$ $[∵y = x]$

$\Rightarrow 2x^2 = 32$

$\Rightarrow x = \pm 4$

So, the intersection point of circle $x^2 + y^2 = 32$ and line $y = x$ are $(4, 4)$ or $(-4, -4)$.

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

Since, $y = 0$

$∴$ $x^2 + (0)^2 = 32$

$\Rightarrow$ $x = \pm 4\sqrt{2}$

So, the circle intersects the X-axis at $(\pm 4\sqrt{2}, 0)$.

As we have to find the area of the region in first quadrant we will take only positive value of $x$ and $y$.

$\text{Area of shaded region} = \int_{0}^{4} x dx + \int_{4}^{4\sqrt{2}} \sqrt{(4\sqrt{2})^2 - x^2} dx$

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

$= \frac{16}{2} + \left[ \frac{4\sqrt{2}}{2} \cdot 0 + 16 \sin^{-1} \frac{(4\sqrt{2})}{(4\sqrt{2})} - \frac{4}{2} \sqrt{(4\sqrt{2})^2 - 16} - 16 \sin^{-1} \frac{4}{4\sqrt{2}} \right]$

$= 8 + \left[ 16 \cdot \frac{\pi}{2} - 2 \cdot \sqrt{16} - 16 \cdot \frac{\pi}{4} \right]$

$= 8 + [8\pi - 8 - 4\pi] = 4\pi \text{ sq. units}$