Question

The area between $x=y^2$ and $x=4$ is divided into two equal parts by the line $x=a$, find the value of $a$.

Answer 1

The line, $x=a$, divides the area bounded by the parabola and $x=4$ into two equal parts.
$\therefore \text{Area}OAD=\text{Area}ABCD$

It can be observed that the given area is symmetrical about x-axis.
$\Rightarrow$ Area $OED$ $=$ Area $EFCD$

Area $OED$ $={\int }_0^{a}ydx$
$={\int }_0^a\sqrt{x}dx$
$={\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^a$
$=\frac{2}{3}(a{)}^{\frac{3}{2}}$ .............. (1)

Area of $EFCD$ $={\int }_a^4\sqrt{x}dx$
$={\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_a^4$

$=\frac{2}{3}[8-a^{\frac{3}{2}}]$ ........... (2)

From (1) and (2), we obtain
$\frac{2}{3}(a{)}^{\frac{3}{2}}=\frac{2}{3}[8-\left(a{)}^{\frac{3}{2}}\right]$
$\Rightarrow 2\cdot (a{)}^{\frac{3}{2}}=8$
$\Rightarrow (a{)}^{\frac{3}{2}}=4$
$\Rightarrow a=(4{)}^{\frac{2}{3}}$

Therefore, the value of a is $(4{)}^{\frac{2}{3}}$.