question_answer

The area of the smaller part of the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\], cut off by the line\[x=\frac{a}{\sqrt{2}}\], is given by

A) \[\frac{{{a}^{2}}}{2}\left( \frac{\pi }{2}+1 \right)\]       

B)        \[\frac{{{a}^{2}}}{2}\left( \frac{\pi }{2}-1 \right)\]

C) \[{{a}^{2}}\left( \frac{\pi }{2}-1 \right)\]       

D)        None of these

Correct Answer: B

Solution :

[b] Area \[=2\int\limits_{a/\sqrt{2}}^{a}{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}\] \[=2{{\left( \frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a} \right)}^{a}}\] \[=2\left\{ \left( 0+\frac{{{a}^{2}}}{2}.\frac{\pi }{2} \right)-\frac{a}{2\sqrt{2}}-\frac{a}{\sqrt{2}}+\frac{a}{2}.\frac{\pi }{4} \right\}\] \[=\frac{{{a}^{2}}}{2}\left( \frac{\pi }{2}-1 \right)\]