Question

Find the common area (in sq. units) enclosed by the parabolas $4y^2=9x$ and $3x^2=16y$.

Answer 1

Red graph indicates the equation $4y^2=9x$ & blue curve indicates$3x^2=16y$
Solving the equation
$4y^2=9x$ ...(1)
and $3x^2=16y$ ...(2)
$\Rightarrow x=\frac{4}{9}{y}^2$
Substitute the value of $x$ in equation (2), we get
$3\left(\frac{16y^4}{81}\right)=16y$ ($÷$ by $16$)
$\Rightarrow 3\left(\frac{y^4}{81}\right)=y$
$\Rightarrow 3y^4-81y=0$
$\Rightarrow 3y(y^3-27)=0$
When $y=0,x=0$
When $y=3,x=4$
$\therefore$ the point of intersection are $(0,0)$ and $(4,3)$.
Required area $={\int }_0^4(y_1-y_2)dx$
$={\int }_0^4(\frac{3}{2}{x}^{1/2}-\frac{3}{16}{x}^3)dx$
$={\int }_0^4{[\frac{3x^{3/2}}{2\frac{3}{2}}-3\frac{x^3}{3\left(16\right)}]}_0^4$
$={[x^{3/2}-\frac{x^3}{\left(16\right)}]}_0^4$
$=(8-4)-0$
$=4$ sq.units