Find the area of the region in the first quadrant enclosed by the x-a…

The correct answer is Option (1) → $4\pi$

The given equations are

$y = x \quad \dots (1)$

and

$x^2 + y^2 = 32 \quad \dots (2)$

Solving $(1)$ and $(2)$, we find that the line and the circle meet at $B(4, 4)$ in the first quadrant. Draw perpendicular $BM$ to the $x$-axis.

Therefore, the required area = area of the region $OBMO$ + area of the region $BMAB$.

Now, the area of the region $OBMO$

$= \int\limits_{0}^{4} y \, dx = \int_{0}^{4} x \, dx \quad \dots (3)$

$= \frac{1}{2} \left[ x^2 \right]_{0}^{4} = 8$

Again, the area of the region $BMAB$

$= \int\limits_{4}^{4\sqrt{2}} y \, dx = \int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx$

$= \left[ \frac{1}{2} x \sqrt{32 - x^2} + \frac{1}{2} \times 32 \times \sin^{-1} \frac{x}{4\sqrt{2}} \right]_{4}^{4\sqrt{2}}$

$= \left( \frac{1}{2} 4\sqrt{2} \times 0 + \frac{1}{2} \times 32 \times \sin^{-1} 1 \right) - \left( \frac{4}{2} \sqrt{32 - 16} + \frac{1}{2} \times 32 \times \sin^{-1} \frac{1}{\sqrt{2}} \right)$

$= 8\pi - (8 + 4\pi) = 4\pi - 8 \quad \dots (4)$

Adding (3) and (4), we get, the required area $= 4\pi$.