Area of the region in the first quadrant enclosed by the $x$ -axis, the line $y = x$ and the circle ${x^2} + {y^2} = 32$ is

• $16\pi$ sq units

• $4\pi$ sq units

• $32\pi$ sq units

• 24 sq units

Correct Option (b)

Point of intersection, $x = 4$

Area of shaded region $= \int_0^4 x dx + \int_4^{4\sqrt 2 } {\sqrt {32 - {x^2}} } dx$

$= \left[ {\frac{{{x^2}}}{2}} \right]_0^4 + \left[ {\frac{x}{2}\sqrt {32 - {x^2}} + 16{{\sin }^{ - 1}}\frac{x}{{4\sqrt 2 }}} \right]_4^{4\sqrt 2 }$

$= 8 + 16\frac{\pi }{2} - 8 - 4\pi = 4\pi Sq.$ units.