The area in the first quadrant between `x^2+y^2=pi^2` and `y=sinx` is

Options

A. `(pi^(3)-8)/(4)` sq units
B. `(pi^(3))/(4)` sq units
C. `(pi^(3)-16)/(4)` sq units
D. `(pi^(3)-8)/(2)` sq units

Answer

A

Explanation

Given curve `x^(2)+y^(2)=pi^(2)` is a circle of radius `pi` and centre at origin.

`therefore` Required area = Area of circle (1st quadrant)
`(pi.pi^(2))/(4)-int_(0)^(pi)sinxdx=(pi.pi^(2))/(4)-[-cosx]_(0)^(pi)`
`=(pi^(3))/(4)+(cospi-cos0)`
`=(pi^(3))/(4)+(-1-1)=(pi^(3))/(4)-2=(pi^(3)-8)/(4)`