# 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 A. 16π sq units B. 4π sq units C. 32π sq units D. 24 sq units

B)

The x-axis, the line y = x and the circle $x^2 + y^2 = 32$ .

$\\ \text{By substitution}; \Rightarrow x^{2}+x^{2}=32\\ \therefore x=4\\ \text{Radius of the circle} =\sqrt{32}=4 \sqrt{2}\\ \text{Required area}\\ =\int_{0}^{4} \mathrm{xdx}+\int_{4}^{4 \sqrt{2}}\left(\sqrt{32-\mathrm{x}^{2}}\right) \mathrm{dx}\\ \left[\int \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}=\frac{\mathrm{x} \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}{2}+\frac{\mathrm{a}^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]$

\begin{aligned} &=\int_{0}^{4} x d x+\int_{4}^{4 \sqrt{2}}\left(\sqrt{(4 \sqrt{2})^{2}-x^{2}}\right) d x\\ &=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{4}+\left[\frac{\mathrm{x} \sqrt{(4 \sqrt{2})^{2}-\mathrm{x}^{2}}}{2}+\frac{(4 \sqrt{2})^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{4 \sqrt{2}}\right)\right]_{4}^{4 \sqrt{2}}\\ &=\left(8-0+\left(\frac{32}{2} \times \frac{\pi}{2}\right)-\frac{16}{2}-\left(\frac{32}{2} \times \frac{\pi}{4}\right)\right)\\ &=(8+8 \pi-8-4 \pi)\\ &=4 \pi \text { sq.units } \end{aligned}