Question

Find the area of the smaller part of the circle $x^2+y^2=a^2$ cut off by the line $x=\frac{a}{\sqrt{2}}$

Answer 1

Given:

$x^2+y^2=a^2$

$x=\frac{a}{\sqrt{2}}$

$\therefore$ Radius $r=a$

Centre $=(0,0)$

Draw diagram
$x^2+y^2=a^2$

$\Rightarrow y^2=a^2-x^2$

$\Rightarrow y=\pm \sqrt{a^2-x^2}$

$\therefore y=\sqrt{a^2-x^2}$

$\text{Area}APQ=2\times \text{Area}APRA$

$=2\times {\int }_{\frac{a}{\sqrt{2}}}^a\sqrt{a^2-x^2}dx$

$=2{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}]}_{\frac{a}{\sqrt{2}}}^a$

$=2[0+\frac{a^2}{2}{\sin }^{-1}(1)-\frac{a}{2\sqrt{2}}\sqrt{\frac{a^2}{2}}-\frac{a^2}{2}{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)]$

$=2[\frac{a^2}{2}\left({\sin }^{-1}\right(1)-{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right))-\frac{a}{2\sqrt{2}}\times \frac{a}{\sqrt{2}}]$

$=2[\frac{a^2}{2}(\frac{\pi }{2}-\frac{\pi }{4})-\frac{a^2}{4}]$
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$=2[\frac{a^2}{2}\left(\frac{\pi }{4}\right)-\frac{a^2}{4}]$

$=2\times \frac{a^2}{4}[\frac{\pi }{2}-1]$
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$=\frac{a^2}{2}[\frac{\pi }{2}-1]$

$\therefore$ Required Area ​$=\frac{a^2}{2}[\frac{\pi }{2}-1]\text{square units}$.

Final answer:
Therefore, required area ​$=\frac{a^2}{2}[\frac{\pi }{2}-1]\text{square units}$.