Question

The area covered inside the parabola $5x^2-y=0$ but outside the parabola $2x^2-y+9=0$ is

• A. $12\sqrt{3}$ sq. units
• B. $6\sqrt{3}$ sq. units
• C. $8\sqrt{3}$ sq. units
• D. $9\sqrt{3}$ sq. units

Answer 1

The correct option is A

$12\sqrt{3}$ sq. units

$y_1=5x^2$
$y_2=2x^2+9$
$A(-\sqrt{3},15);B(\sqrt{3},15)$
Since the curve is symmetric about the $y-$axis
Area:
$=2{\int }_0^{\sqrt{3}}(2x^2+9-5x^2)dx=2{[9x-x^3]}_0^{\sqrt{3}}$
$=2(9\sqrt{3}-3\sqrt{3})=12\sqrt{3}$ sq. units