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

and circle, $x^2+y^2=a^2$ ...(iii)

Since, given line cuts the circles so put $x=\frac{a}{\sqrt{2}}$ in Eq. (ii), we get

$(\frac{a}{\sqrt{2}}{)}^2+y^2=a^2\Rightarrow \Rightarrow y^2=\frac{a^2}{1}-\frac{a^2}{2}=\frac{a^2}{2}\Rightarrow |y|=\frac{a}{\sqrt{2}}\Rightarrow y=\pm \frac{a}{\sqrt{2}}Wheny=\frac{a}{{\sqrt{2}}^{\prime }}{x}^2=a^2-(\frac{a^2}{2})\Rightarrow x^2=\frac{a^2}{2}\Rightarrow x=\pm \frac{a}{\sqrt{2}}$
$\therefore$ Intersection point in first quadrant is $(\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}})$

Required area = 2(Area of shaded region in first quadrant only)
$=2{\int }_{\frac{a}{\sqrt{2}}}^a|y|dx=2{\int }_{\frac{a}{\sqrt{2}}}^a\sqrt{a^2-x^2}dx=2[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}(\frac{x}{a}){]}_{\frac{a}{\sqrt{2}}}^a$

$[\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}(\frac{x}{a}\left)\right]$
$=2[0+\frac{a^2}{2}si{n}^{-1}(1)-\frac{a}{2\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}}-\frac{a^2}{2}si{n}^{-1}(\frac{1}{\sqrt{2}}\left)\right]=2\left[\frac{a^2}{2}\right(\frac{\pi }{2})-\frac{a}{2\sqrt{2}}.\frac{a}{\sqrt{2}}(\frac{\pi }{4}\left)\right]=2[\frac{a^2\pi }{4}-\frac{\pi a^2}{8}-\frac{a^2}{4}]=(\frac{a^2\pi }{4}-\frac{a^2}{2})=\frac{a^2}{2}(\frac{\pi }{2}-1)squnit$
Therefore, the area of smaller part of the circle, $x^2+y^2=a^2,cutoffbytheline,x=\frac{a}{\sqrt{2}},is\frac{a^2}{2}(\frac{\pi }{2}-1)squnit$