Question

Area bounded by $y=2\sqrt{x}$ and $x=3\sqrt{y}$ is equal to (in sq. units)

• A. 12
• B. 8
• C. 10
• D. 6

Answer 1

The correct option is A 12

First we have to find the intersection point.
Hence, $2\sqrt{x}={\left(\frac{x}{3}\right)}^2$
$\Rightarrow 18\sqrt{x}=x^2$
$\Rightarrow \sqrt{x}(18-x\sqrt{x})=0$

$\Rightarrow x=0$ and $x\sqrt{x}=18$
$\Rightarrow x^3={18}^2$
Now the area bounded by the curve is
$I={\int }_0^{{18}^{\frac{2}{3}}}2\sqrt{x}-\frac{x^2}{9}dx$
$={[\frac{4x^{\frac{3}{2}}}{3}-\frac{x^3}{27}]}_0^{{18}^{\frac{2}{3}}}$
$=\frac{4\times 18}{3}-\frac{{18}^2}{27}$
$=24-12$
$=12$ sq units.