Mathematics - Area Under Curves Question with Solution | TestHub

The given curves are $y=a{x}^2$ and $x=a{y}^2$.

To find the points of intersection, substitute $y=a{x}^2$ into $x=a{y}^2$:

$x=a(a{x}^2{)}^2$

$x=a^3{x}^4$

$x(a^3{x}^3-1)=0$

Since $a>0$, the points of intersection are $x=0$ and $x=\frac{1}{a}$.

The area bounded by the curves is given by:

$A={\int }_0^{1/a}\left(\sqrt{\frac{x}{a}}-a{x}^2\right)dx$

$A={\left[\frac{1}{\sqrt{a}}\frac{x^{3/2}}{3/2}-a\frac{x^3}{3}\right]}_0^{1/a}$

$A=\frac{2}{3\sqrt{a}}{\left(\frac{1}{a}\right)}^{3/2}-\frac{a}{3}{\left(\frac{1}{a}\right)}^3$

$A=\frac{2}{3a^2}-\frac{1}{3a^2}=\frac{1}{3a^2}$

Given that the area is $1$ sq. unit:

$\frac{1}{3a^2}=1$

$a^2=\frac{1}{3}$

Since $a>0$, we get $a=\frac{1}{\sqrt{3}}$.

Answer: $\frac{1}{\sqrt{3}}$