Find the area of the smaller part of the circle `x^2+y^2=a^2` cut off by the line `x=a/(sqrt(2))`

Explanation

`therefore` Area `{ABCD}=2 times` Area `{ABC}`


Area of ` A B C =int_{sqrt{2}}^{x} y d x =int_{frac{1}{sqrt{2}}}^{n} sqrt{a^{2}-x^{2}} d x `


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


` =[frac{a^{2}}{2}(frac{pi}{2})-frac{a}{2 sqrt{2}} sqrt{.a^{2}-frac{a^{2}}{2}-frac{a^{2}}{2} sin ^{-1}(frac{1}{sqrt{2}})]}. `


`=frac{a^{2} pi}{4}-frac{a}{2 sqrt{2}} cdot frac{a}{sqrt{2}}-frac{a^{2}}{2}(frac{pi}{4})`


`=frac{a^{2} pi}{4}-frac{a^{2}}{4}-frac{a^{2} pi}{8} =frac{a^{2}}{4}[pi-1-frac{pi}{2}]`


`=frac{a^{2}}{4}[frac{pi}{2}-1] Rightarrow text { Area } A B C D`


`=2[frac{a^{2}}{4}(frac{pi}{2}-1)]=frac{a^{2}}{2}(frac{pi}{2}-1) `


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