# Find the common area enclosed by the parabolas $4y^{2}=9x$ and $3x^{2}=16y$

Toolbox:
• Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx$ or $\int \limits _a^b y dx$
• If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Solving $3x^2 =16 y, 4y^2 =9x$
$4 \bigg(\large\frac{3x^2}{16}\bigg)^2 $$=9x \large\frac{36x^4}{256}$$=9x$
$x^4= 64x$
$x(x^3-4^3)=0=>x=0,4$
Required area= Area $4y^2=9x$ between $x=0\;x=4$ -area $3x^2=16 y$ between $x=0,x=4$