Question

Find the area enclosed between the parabola $y^2=z$ and $x^2=y$.

Answer 1

$x^2=y$ and $y^2=x$ are two parabola

$y^2=a$ $\Rightarrow y=\sqrt{x}$

$\therefore$ We need to find the area enclosed between the functions

$y_1=x^2$ and $y_2=\sqrt{x}$

$y_1=y_2\Rightarrow x^2=x^{\frac{1}{2}}\Rightarrow x^2-x^{\frac{1}{2}}=0\Rightarrow x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)=0$

Thus intersection of the two functions occurred $x=0$ and $x=!$

When $x=(0,10)$ the function $y_2\ge y_1$, and that therefore the difference between the two functions is given by the function $2\in IR$ where $z=x^{\frac{1}{2}}-x^2$

$z=\int zdz=\frac{2}{3}{x}^{\frac{3}{2}}-\frac{1}{3}{x}^3+C$

${\int }_0^{1}zdz=\left\{\frac{2}{3}\right(1{)}^{\frac{3}{2}}-\left(\frac{1}{3}\right)(1{)}^3+c\}-\left\{\frac{2}{3}\right(0{)}^{\frac{2}{3}}-\left(\frac{1}{3}\right)\left(0\right)\}$

$=\frac{2}{3}-\frac{1}{3}+C-0+0-C=\frac{1}{3}$

The area enclosed between the parabola $x^2=y$ and $y^2=x$ is $\frac{1}{3}$.