Answer
$\dfrac{8\pi}{3}$
Work Step by Step
Our aim is to compute the volume of the given curve when it is revolved around $y$-axis by using the shell method.
Shell method for computing the volume of the curve: Let us consider two functions $f(x)$ and $g(x)$ (both are continuous functions) with $f(x) \geq g(x)$ on the interval $[m, n]$ and $R$ defines the region which revolves about the y-axis and is known as the region bounded by the curves $y=f(x)$ and $y=g(x)$ between the lines $x=m$ and $x=n$. Then, the volume of the solid can be expressed as:
$Volume, V=\int_m^n 2 \pi x [f(x)-g(x)] \ dx$
Here, $2\pi x=\text{Shell Circumference}$ and $[f(x)-g(x)] =\text {Height of the shell}$
We are given that $y=\sqrt{4-2x^2}; y=0$ on $[0, \sqrt {2}]$
Thus, $V=\int_m^n 2 \pi x [f(x)-g(x)] \ dx=2 \pi \int_0^{\sqrt {2}} x [\sqrt{4-2x^2}] \ dx$
Suppose that $t=4-2x^2 \implies da=-4x \ dx$
Now, $V=2 \pi \int_4^{0} (\sqrt a) \times \dfrac{-\ da}{4}$
or, $= - \dfrac{2 \pi}{4}[\dfrac{t^{1/2}}{1/2}]_4^0 $
or, $=\dfrac{8\pi}{3}$