Definite Integration Question 195
Question: What is the area enclosed by the equation $ x^{2}+y^{2}=2 $ -
Options:
A) $ 4\pi $ square units
B) $ 2\pi $ square units
C) $ 4{{\pi }^{2}} $ square units
D) 4 square units
Show Answer
Answer:
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[ \frac{x}{2}\sqrt{2-x^{2}}+\frac{2}{2}{{\sin }^{-1}}\frac{x}{\sqrt{2}} ]_0^{\sqrt{2}} $
$ =4[ {{\sin }^{-1}}\frac{\sqrt{2}}{\sqrt{2}} ]=4{{\sin }^{-1}}1=4\times \frac{\pi }{2}=2\pi sq.unit. $
BETA
