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Q.
The area bounded by $y=\left(2x\right)^{1 /2}$ and $x=\left(2y\right)^{1 /2}$ is
Application of Integrals
Solution:
We have, $y=\sqrt{2x}$
$\Rightarrow y^{2}=2x \, \cdots\left(i\right)$ a parabola with vertex $\left(0,0\right)$.
and $x = \left(2y\right)^{1 /2}$
$\Rightarrow \, x^{2}=2y \, \ldots\left(ii\right)$ a parabola with vertex $\left(0,0\right)$.
$\left(i\right)$ and $\left(ii\right)$ intersects at $\left(0, 0\right)$ and $\left(2, 2\right)$.
Required area = area of shaded region
$=\int\limits_{0}^{2}\left(\sqrt{2x}-\frac{x^{2}}{2}\right) dx$
$=\frac{2}{3} \sqrt{2} \left|x^{3/ 2}\right|_{0}^{2}-\frac{1}{2}\left|\frac{x^{3}}{3}\right|_{0}^{2}$
$=\frac{2\sqrt{2}}{3}\left(2\sqrt{2}\right)-\frac{1}{2}\times\frac{8}{3}$
$=\frac{8}{3}-\frac{4}{3}$
$=\frac{4}{3}$ sq. units
