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Find the area of the smaller part of th...

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))`

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The correct Answer is:
`(a^(2))/(2)((pi)/(2)-1)" sq. units"`
The area of the smaller part of the circle, `x^(2)+y^(2)=a^(2)`, cut off by the line, `x=(a)/(sqrt(2))`, is the area ABCDA.

Solving `x^(2)+y^(2)=a^(2) and x=(a)/(sqrt(2))` for their points of intersection, we get
`(a^(2))/(2)+y^(2)=a^(2)`
`"or "y^(2)=(a^(2))/(2)or y=pm(a)/(sqrt(2))`
`therefore" Area ABCD=2"xx"Area ABC"`
`=2overset(a)underset((a)/(sqrt(2)))intsqrt(a^(2)-x^(2))dx=2[(x)/(2)sqrt(a^(2)-x^(2))+(a^(2))/(2)sin^(-1)""(x)/(a)]_((a)/(sqrt(2)))^(a)`
`=2[(a^(2))/(2)((pi)/(2))-(a)/(2sqrt(2))sqrt(a^(2)-(a^(2))/(2)-(a^(2))/(2))sin^(-1)((1)/(sqrt(2)))]`
`=2[(a^(2)pi)/(4)-(a^(2))/(4)-(a^(2)pi)/(8)]`
`=(a^(2))/(2)((pi)/(2)-1)` sq. units.
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