Question

Find the area of the region bounded by curves $y^2=4x,x^2=4y$

Answer 1

We have $y^2=4x\cdots \left(1\right)$
$x^2=4y\cdots \left(2\right)$

Substitute the ralue of $y$ from equation (2) to equation (1)
$\Rightarrow \frac{x^4}{16}-4x=0$
$\Rightarrow x(x^3-64)=0$
$\Rightarrow x=0,4$

when $x=0\Rightarrow y=0$and

when $x=4\Rightarrow y=4$ [using (2)]

Hence $(0,0)\text{and}(4,4)$are the points of intersection.

So, the area bounded by curves is shaded in the diagram below:

Area $={\int }_{x_1}^{x_2}(y_2-y_1)dx$
$\Rightarrow \text{Area}={\int }_0^4(2\sqrt{x}-\frac{x^2}{4})dx$
$[\because x\text{varies from}0\text{to}4]$

$\Rightarrow \text{Area}={[2\times \frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^3}{12}]}_0^4$
$[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$

$\Rightarrow \text{Area}=\frac{32}{3}-\frac{64}{12}$

$\Rightarrow \text{Area}=\frac{32\times 4-64}{12}$

$\Rightarrow \text{Area}=\frac{16}{3}sq.units.$

Hence,the required area is $\frac{16}{3}sq.units.$