question_answer

What is the area enclosed by the equation\[{{x}^{2}}+{{y}^{2}}=2\]?

A) \[4\pi \] square units

B) \[2\pi \] square units

C) \[4{{\pi }^{2}}\] square units

D) 4 square units

Correct Answer: B

Solution :

[b] Given equation of circle is \[{{x}^{2}}+{{y}^{2}}=2\] \[\Rightarrow y=\sqrt{2-{{x}^{2}}}\] Required area \[=4\times \] Area of shaded portion \[=4\int_{0}^{\sqrt{2}}{\sqrt{2-{{x}^{2}}}dx}\] \[=4\left[ \frac{x}{2}\sqrt{2-{{x}^{2}}}+\frac{2}{2}{{\sin }^{-1}}\frac{x}{\sqrt{2}} \right]_{0}^{\sqrt{2}}\] \[=4\left[ {{\sin }^{-1}}\frac{\sqrt{2}}{\sqrt{2}} \right]=4{{\sin }^{-1}}1=4\times \frac{\pi }{2}=2\pi \,sq.\,\,unit.\]