Question

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. $\frac{a^2}{2}(\frac{\pi }{2}+1)$
• B. $\frac{a^2}{2}(\frac{\pi }{2}-1)$
• C. $a^2(\frac{\pi }{2}-1)$
• D. None of these

Answer 1

Answer: (B)

$\frac{a^2}{2}(\frac{\pi }{2}-1)$

Required area $2{\int }_{\frac{a}{\sqrt{2}}}^{a}ydx=2{\int }_{\frac{a}{\sqrt{2}}}^a\sqrt{a^2-x^2}dx$
$=2{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}]}_{\frac{a}{\sqrt{2}}}^a$
$=2[\frac{a^2}{2}.\frac{\pi }{2}-(\frac{a}{2\sqrt{2}}.\frac{a}{\sqrt{2}}+\frac{a^2}{2}.\frac{\pi }{4})]$
$=\frac{a^2}{2}(\frac{\pi }{2}-1)$