The ellipse $4x^2+y^2=8$ has semi-axes $a_x=\sqrt{2}$, $a_y=2\sqrt{2}$. The parabola $y^2=4x$ opens rightward.

Intersection: $4x+4x^2=8\Rightarrow x^2+x-2=0\Rightarrow x=1$ (taking positive root). So they meet at $(1,\pm 2)$.

By symmetry about the $x$-axis, the area is twice the upper half. Upper region splits into:

(i) Under parabola from $x=0$ to $1$: ${\int }_0^{1}2\sqrt{x}dx=\frac{4}{3}$.

(ii) Under ellipse from $x=1$ to $\sqrt{2}$: ${\int }_1^{\sqrt{2}}\sqrt{8-4x^2}dx=2{\int }_1^{\sqrt{2}}\sqrt{2-x^2}dx$.

Using $\int \sqrt{2-x^2}dx=\frac{x\sqrt{2-x^2}}{2}+{\sin }^{-1}\left(\frac{x}{\sqrt{2}}\right)$:

$2{\left[\frac{x\sqrt{2-x^2}}{2}+{\sin }^{-1}\frac{x}{\sqrt{2}}\right]}_1^{\sqrt{2}}=2\left[\frac{\pi }{2}-\frac{1}{2}-\frac{\pi }{4}\right]=\frac{\pi }{2}-1$.

Total area $=2\left[\frac{4}{3}+\frac{\pi }{2}-1\right]=\frac{8}{3}+\pi -2=\pi +\frac{2}{3}$.