The area of the region bounded by $x=y^2$ and the line $x=4$ in square units is :

$\frac{32}{3}$

$x=y^2, ~~x=4$

for $x=4$

$y^2=4$

$y= \pm 2$

point of Intersection of $x=4~~~y^2=x$ are $(4,2)$ and $(4,-2)$

cercus is symmetrical about x axis

area of port I = area of port II

area of figure (Ar.) = 2 × area of port I

Ar. = $2 x \int\limits_0^4 \sqrt{x} d x$

$x=y^2$

$\Rightarrow y=\sqrt{x}$

Ar. = $2 \int\limits_0^4 x^{1 / 2} d x =2\left[\frac{x^{1 / 2+1}}{\frac{1}{2}+1}\right]_0^4=2\left[\frac{x^{3 / 2}}{\frac{3}{2}}\right]_0^4$

$=\frac{2 \times 2}{3} \times 4^{3 / 2}-0$

Ar. = $\frac{4}{3} \times 8=\frac{32}{3}$