Question

The area enclosed by the circle x² + y² = 2 is equal to
(a) 4π sq. units
(b) $2\sqrt{2}\mathrm{\pi }$sq. units
(c) 4π² sq. units
(d) 2π sq. units

Answer 1

To find: area enclosed by the circle x² + y² = 2



The whole area enclosed by the given circle = 4(Area of the region bounded by AOBA)

Thus,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}=4{\int }_0^{\sqrt{2}}ydx \\ =4{\int }_0^{\sqrt{2}}\left(\sqrt{2-x^2}\right)dx \\ =4{\left(\frac{x}{2}\sqrt{2-x^2}+\frac{{\left(\sqrt{2}\right)}^2}{2}{\sin }^{-1}\frac{x}{\sqrt{2}}\right)}_0^{\sqrt{2}} \\ =4\left[\left(\frac{\sqrt{2}}{2}\sqrt{2-{\sqrt{2}}^2}+\frac{{\left(\sqrt{2}\right)}^2}{2}{\sin }^{-1}\frac{\sqrt{2}}{\sqrt{2}}\right)-\left(\frac{0}{2}\sqrt{2-0^2}+\frac{{\left(\sqrt{2}\right)}^2}{2}{\sin }^{-1}\frac{0}{\sqrt{2}}\right)\right] \\ =4\left[\left(\frac{\sqrt{2}}{2}\sqrt{2-2}+\frac{2}{2}{\sin }^{-1}1\right)-\left(0+\frac{2}{2}{\sin }^{-1}0\right)\right] \\ =4\left[\left(0+\frac{\mathrm{\pi }}{2}\right)-\left(0\right)\right] \\ =4\left(\frac{\mathrm{\pi }}{2}\right) \\ =2\mathrm{\pi } \\ \mathrm{Thus},\mathrm{Area}=2\mathrm{\pi }\mathrm{sq}.\mathrm{units} \end{gathered}$$


Hence, the correct option is (d).