$xy= 16-\gt$ is a hyperbola $x=y$ is also a line $x=8$ straight line Intersection of line $x=y$ and hyperbola $xy=16$

put $x=y$ in equation $xy=16.$

$\therefore x^2=16$

$\therefore x=4$

When $x=4 ,y=4 $

$\therefore $ point of intersection is $(4,4)$

Now the region is divided into two parts since upper limit changes. At $x=4.$

Region 1 i.e. $\triangle AOD$

1) Outer limit x

$X=0$ to $x=4 $

2) Inner limit y

a) Upper limit $y=x$

b) Lower limit $y=0$

Region 2: i.e. $\triangle BCD$

1) Outer limit x

$X=4$ to $x=8 $

2) Inner limit y.

a) upper limit is equation of hyperbola.

i.e. $xy =16$

$\therefore y=16/x$

b) Lower limit $y=0$

Integration becomes

$$I=\int\limits_{x=0}^{x=4}\int\limits_{y=0}^{y=x}x^2dydx+\int\limits_{x=4}^{x=8}\int\limits_{y=0}^{y=\frac {16}x}x^2\space dydx$$

$I=\int\limits_{x=0}^{x=4}x^2[y]^x_0dx+\int\limits_{x=4}^{x=8}x^2[y]^{\frac {16}x}_0 dx\\ I=\int\limits^4_{x=0}x^3dx+\int\limits_{x=4}^{x=8}x^2\Bigg(\dfrac {16}x\Bigg)dx\\ =\Bigg[\dfrac {x^4}4\Bigg]^4_0+16 \int\limits_4^8xdx\\ =\dfrac {x^4}4+16\Bigg[\dfrac {x^2}2\Bigg]^8_4\\ =4^3+\dfrac {16}2[8^2-4^2]\\ =64+ 8[48]\\ =448$