Find the area of the region in the first quadrant enclosed by the x-axis, the line `y" "=" x"` , and the circle `x^2+y^2=32` .

Explanation

`x^(2)+y^(2)=4`
`impliesy=sqrt(2^(2)-x^(2)) " " `...(1)
`y-x " " ` ...(2)
Their points of intersection are `Q(-sqrt(2), -sqrt(2))` and `P(sqrt(2), sqrt(2)).`

In first quardrant, PB `bot` OX.
The centre adn radius of the circle are (0, 0) and 2 units respectively.
Now the required area
`=` Area of OBP + Area of BAP
`=int_(0)^(sqrt(2))xdx+int_(sqrt(2))^(2)sqrt(2^(2)-x^(2))dx`
`=[(x^(2))/(2)]_(0)^(sqrt(2))+[(x)/(2)sqrt(2^(2)-x^(2))+(2^(2))/(2)"sin"^(-1)(x)/(2)]_(sqrt(2))^(2)`
`=(1-0)+[(0+2"sin"^(-1))-((sqrt(2))/(2).sqrt(2)+2"sin"^(-1)(1)/(sqrt(2)))]`
` =1+2.(pi)/(2)-1-2.(pi)/(4)=(pi)/(2)sq.` units.