Question

The area bounded by $y^2=4x$ and $x^2=4y$ is

• A. $\frac{20}{3}$ sq. units
• B. $\frac{16}{3}$ sq. units
• C. $\frac{14}{3}$ sq. units
• D. $\frac{10}{3}$ sq. units

Answer 1

The correct option is B $\frac{16}{3}$ sq. units

The intersection point of a curves $y^2=4x$ and $x^2=4y$ are $O(0,0)$ and $A(4,4)$
$\therefore$ Required area ${\int }_0^4(y_2-y_1)dx$
$={\int }_0^4(\sqrt{4x}-\frac{x^2}{4})dx$
$={[\frac{2x^{3/2}}{3/2}-\frac{x^3}{12}]}_0^4$
$=\frac{4}{3}[4^{3/2}-0]-\frac{1}{12}[4^3-0]$
$=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$sq. units