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))`

Options

A. `(a^(2))/(2)((pi)/(2) + 1)`
B. `(a^(2))/(2)((pi)/(2) - 1)`
C. `(a^(2))((pi)/(2) - 1)`
D. None of these

Answer

B

Explanation

Area = `2 underset(a//sqrt2)overset(a)(int)sqrt(a^(2) - x^(2)) dx `
=`2((x)/(2) sqrt(a^(2) - x^(2)) + (a^(2))/(2)sin^(-1)(x)/(a))^(a)`
`= 2{(0 + (a^(2))/(2) .(pi)/(2))- (a)/(2sqrt2). (a)/(sqrt2)+ (a^(2))/(2) . (pi)/(4)}`
`= (a^(2))/(2) ((pi)/(2) - 1)`