JEE 2023 — Mathematics PYQ

The given curves are  
$x^2\le y,y\le 8-x^2,y\le 7$.

On solving, we get  

$x^2=8-x^2$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

So, area $=2\left[{\int }_0^4\sqrt{y}dy<br>+{\int }_4^7\sqrt{8-y}dy\right]$

$=2\left[{\frac{y^{3/2}}{3/2}|}_0^4<br>+{-\frac{(8-y{)}^{3/2}}{3/2}|}_4^7\right]$

$=2\times \frac{2}{3}<br>\left[4^{3/2}-0+(-1{)}^{3/2}+4^{3/2}\right]$

$=\frac{4}{3}\{8-1+8\}<br>=\frac{4}{3}\times 15<br>=20\text{sq. units}$