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

Options

A. `16pi" sq units"`
B. `4pi" sq units"`
C. `32pi" sq units"`
D. `24 pi " sq units"`

Answer

B

Explanation

We have, area enclosed by X-axis i.e., y = 0, y = x and the circle `x^(2)+y^(2)=32` in first quadrent.
Since, `x^(2)+(x)^(2)=32 [ :' y=x]`
`rArr 2^(2)=32`
`rArr x=pm4`
So, the intersection point of circle `x^(2)+y^(2)=32" and line "y=x" are "(4,4)" or "(-4,4)`.
and `x^(2)+y^(2)=(4sqrt2)^(2)`
Since, `y=0`
`:. x^(2)+(0)^(2)=32`
`rArr x=pm4sqrt2`
So, the circle intersects the X-axis at `(pm4sqrt2, 0)`.

Area of shaded region `=underset(0)overset(4)intxdx+underset(4)overset(4sqrt2)intsqrt((4sqrt2)^(2)-x^(2))dx`
`=abs(x^(2)/2)_(0)^(4)+abs(x/2sqrt((4sqrt(2))^(2)-x^(2))+((4sqrt2)^(2))/2sin^(-1).x/(4sqrt2))_(4)^(4sqrt2)`
`=16/2+[(4sqrt2)/2 .0+16sin^(-1).((4sqrt2))/((4sqrt2))-4/2sqrt((4sqrt2)^(2)-16)-16sin^(-1). 4/(4sqrt2)]`
`=8+[16. pi/2-2.sqrt16-16. pi/4]`
`=8+[8pi-8-4pi]=4pi" sq units"`