The area (in square units) bounded by the curves $y^2=4 x$ and $x^2=4 y$ in the plane is

Question

Aicharya · Accepted Answer

Given curves are

The intersecting points of Eqs. (i) and (ii) are $(0,0)$ and $(4,4)$

$\therefore$ Required area $=$ Area of shaded portion $O A B C$ $\begin{aligned} & =\int_0^4\left(\sqrt{4 x}-\frac{x^2}{4}\right) d x \\ & =\left[\frac{2 \cdot x^{3 / 2}}{3 / 2}-\frac{x^3}{12}\right]_0^4 \\ & =\left[\frac{4}{3} \cdot 8-\frac{64}{12}\right]=\left[\frac{32}{3}-\frac{16}{3}\right] \end{aligned}$ $=\frac{16}{3}$ sq. units.

The area (in square units) bounded by the curves $y^2=4 x$ and $x^2=4 y$ in the plane is

Solution