Question
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
Answer :
$$2\pi $$ square units
Solution :
Given equation of circle is $${x^2} + {y^2} = 2 \Rightarrow y = \sqrt {2 - {x^2}} $$
Required area $$ = 4 \times $$ Area of shaded portion
$$\eqalign{ & = 4\int_0^{\sqrt 2 } {\sqrt {2 - {x^2}} } dx \cr & = 4\left[ {\frac{x}{2}\sqrt {2 - {x^2}} + \frac{2}{2}{{\sin }^{ - 1}}\frac{x}{{\sqrt 2 }}} \right]_0^{\sqrt 2 } \cr & = 4\left[ {{{\sin }^{ - 1}}\frac{{\sqrt 2 }}{{\sqrt 2 }}} \right] \cr & = 4\,{\sin ^{ - 1}}1 \cr & = 4 \times \frac{\pi }{2} \cr & = 2\pi {\text{ sq}}{\text{. units}} \cr} $$
Given equation of circle is $${x^2} + {y^2} = 2 \Rightarrow y = \sqrt {2 - {x^2}} $$
Required area $$ = 4 \times $$ Area of shaded portion
$$\eqalign{ & = 4\int_0^{\sqrt 2 } {\sqrt {2 - {x^2}} } dx \cr & = 4\left[ {\frac{x}{2}\sqrt {2 - {x^2}} + \frac{2}{2}{{\sin }^{ - 1}}\frac{x}{{\sqrt 2 }}} \right]_0^{\sqrt 2 } \cr & = 4\left[ {{{\sin }^{ - 1}}\frac{{\sqrt 2 }}{{\sqrt 2 }}} \right] \cr & = 4\,{\sin ^{ - 1}}1 \cr & = 4 \times \frac{\pi }{2} \cr & = 2\pi {\text{ sq}}{\text{. units}} \cr} $$