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

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Solution

## 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, $\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}=4{\int }_{0}^{\sqrt{2}}ydx\phantom{\rule{0ex}{0ex}}=4{\int }_{0}^{\sqrt{2}}\left(\sqrt{2-{x}^{2}}\right)dx\phantom{\rule{0ex}{0ex}}=4{\left(\frac{x}{2}\sqrt{2-{x}^{2}}+\frac{{\left(\sqrt{2}\right)}^{2}}{2}{\mathrm{sin}}^{-1}\frac{x}{\sqrt{2}}\right)}_{0}^{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=4\left[\left(\frac{\sqrt{2}}{2}\sqrt{2-{\sqrt{2}}^{2}}+\frac{{\left(\sqrt{2}\right)}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\sqrt{2}}{\sqrt{2}}\right)-\left(\frac{0}{2}\sqrt{2-{0}^{2}}+\frac{{\left(\sqrt{2}\right)}^{2}}{2}{\mathrm{sin}}^{-1}\frac{0}{\sqrt{2}}\right)\right]\phantom{\rule{0ex}{0ex}}=4\left[\left(\frac{\sqrt{2}}{2}\sqrt{2-2}+\frac{2}{2}{\mathrm{sin}}^{-1}1\right)-\left(0+\frac{2}{2}{\mathrm{sin}}^{-1}0\right)\right]\phantom{\rule{0ex}{0ex}}=4\left[\left(0+\frac{\mathrm{\pi }}{2}\right)-\left(0\right)\right]\phantom{\rule{0ex}{0ex}}=4\left(\frac{\mathrm{\pi }}{2}\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi }\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{Area}=2\mathrm{\pi }\mathrm{sq}.\mathrm{units}$ Hence, the correct option is (d).

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