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

Given curve $x=y^2$ is a parabola symmetrical about X-axis and passing through the origin.

The line x = a, divides the area bounded by the parabola and x = 4 into two equal parts.

Area OAD = Area ABCD

$\therefore$ Area OED = Area EFCD

$\Rightarrow$ Area OED = ${\int }_0^{a}ydx$

and area of $EFCD={\int }_a^4\sqrt{x}dx$ $(\because y^2=x\Rightarrow |y|=\sqrt{x})$

$\Rightarrow {\int }_0^a\sqrt{x}dx={\int }_a^4\sqrt{x}dx\Rightarrow [\frac{x^{\frac{3}{2}}}{\frac{3}{2}}{]}_0^a=[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}{]}_a^4\Rightarrow \frac{2}{3}[a^{\frac{3}{2}}-0]=\frac{2}{3}[4^{\frac{3}{2}}-a^{\frac{3}{2}}]\Rightarrow a^{\frac{3}{2}}=4^{\frac{3}{2}}-a^{\frac{3}{2}}\Rightarrow 2a^{\frac{3}{2}}=8\Rightarrow a^{\frac{3}{2}}=4\Rightarrow a=(4{)}^{\frac{2}{3}}.$
Therefore, the value of a is $(4{)}^{\frac{2}{3}}.$