Question

The area, in square units of the region bounded by the parabolas $y^2=4x$ and $x^2=4y$ is

• A. $\frac{16}{3}$
• B. $\frac{32}{3}$
• C. $\frac{8}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $\frac{16}{3}$

The given parabolas are $y^2=4x$ and $x^2=4y$
The point of intersection of $x^2=4y$ and $y^2=4x$ will be
$x=4$ and $x=0$

Then, the required area will be
$\stackrel{4}{\underset{0}{\int }}(\sqrt{4x}-\frac{x^2}{4})dx$

$={[\sqrt{4}\times \frac{2}{3}{x}^{3/2}-\frac{x^3}{12}]}_0^4$
$=\frac{4\times 8}{3}-\frac{4^3}{12}$

$=\frac{32}{3}-\frac{16}{3}$

$=\frac{16}{3}squnits$