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

Given `x=(a)/(sqrt(2)) " "`...(1)
and circle , `x^(2)+y^(2)=a^(2) " " `...(2)
On solving , `((a)/(sqrt(2)))^(2)+y^(2)=a^(2)`

`implies y^(2)=(a^(2))/(1)-(a^(2))/(2)=(a^(2))/(2)`
`implies y=+-(a)/(sqrt(2))`
` y=(a)/(sqrt(2))impliesx^(2)=a^(2)-((a^(2))/(2))`
`x^(2)=(a^(2))/(2)implies x=+-(a)/(sqrt(2))`
` :. ` In first quadrant, the point of intersection is `((a)/(sqrt(2)), (a)/(sqrt(2))).`
` :. ` Required area ` =2xx`(area of shaded region in first quadrant only)
`=2int_((a)/(sqrt(2)))^(a)sqrt(a^(2)-x^(2))dx`
`=2[(x)/(a)sqrt(a^(2)-x^(2))+(a^(2))/(2)sin^(-1)((x)/(a))]_(a//sqrt(2))^(a)`
`=2[0+(a^(2))/(2)sin^(-1)(1)-(a)/(2sqrt(2))sqrt(a^(2)-(a^(2))/(2))-(a^(2))/(2)sin^(-1)((1)/(sqrt(2)))]`
`=2[(a^(2))/(2)((pi)/(2))-(a)/(2sqrt(2))*(a)/(sqrt(2))-(a^(2))/(2)((pi)/(4))]`
`=2[(a^(2)pi)/(4)-(pia^(2))/(8)-(a^(2))/(4)]=((a^(2)pi)/(4)-(a^(2))/(2))`
` =(a^(2))/(2)((pi)/(2)-1)` sq. unit.