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Solution

From $y^2\le 4x$, we have $x\ge \dfrac{y^2}{4}$. From $4x^2+y^2\le 8$, right boundary is $x=\dfrac12\sqrt{8-y^2}$. Intersection occurs when $4\left(\dfrac{y^2}{4}\right)^2+y^2=8\Rightarrow y^2=4$, so $-2\le y\le 2$. Required area $=\int_{-2}^{2}\left(\dfrac12\sqrt{8-y^2}-\dfrac{y^2}{4}\right)dy = \int_0^2\sqrt{8-y^2}\,dy-\int_0^2\dfrac{y^2}{2}\,dy = (\pi+2)-\dfrac43 = \pi+\dfrac23$.