The area between $x=y^{2}$ and $x=4$ is divided into two equal parts by the line $x=a$, find the value of $a$.
The line, $x=a$, divides the area bounded by the parabola and $x=4$ into two equal parts.
$\therefore$ Area $\mathrm{OAD}=$ Area $\mathrm{ABCD}$
It can be observed that the given area is symmetrical about x-axis.
⇒ Area OED = Area EFCD
Area of EFCD $=\int_{a}^{4} \sqrt{x} d x$
$=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{a}^{4}$
$=\frac{2}{3}\left[8-a^{\frac{3}{2}}\right]$ ...(2)
From (1) and (2), we obtain
$\frac{2}{3}(a)^{\frac{3}{2}}=\frac{2}{3}\left[8-(a)^{\frac{3}{2}}\right]$
$\Rightarrow 2 \cdot(a)^{\frac{3}{2}}=8$
$\Rightarrow(a)^{\frac{3}{2}}=4$
$\Rightarrow a=(4)^{\frac{2}{3}}$
Therefore, the value of $a$ is $(4)^{\frac{2}{3}}$.