Find the volume when the region bounded by the $y=\sqrt{4x}$ and $y=\sqrt{2-x}$ is rotated about the $x$ axis.

Step 1 ：The volume of a solid of revolution can be found using the formula for the volume of a disk or washer. The formula is $\pi {\int }_a^b[R(x){]}^2-[r(x){]}^{2}dx$, where $R(x)$ is the outer radius and $r(x)$ is the inner radius. In this case, the outer radius is $y=\sqrt{4x}$ and the inner radius is $y=\sqrt{2-x}$.

Step 2 ：The limits of integration are the points where the two curves intersect, which can be found by setting the two equations equal to each other and solving for $x$.

Step 3 ：The two curves intersect at $x=0$ and $x=2$. So, the limits of integration are $a=0$ and $b=2$.

Step 4 ：Substitute these values into the volume formula to get the volume of the solid when the region bounded by the $y=\sqrt{4x}$ and $y=\sqrt{2-x}$ is rotated about the $x$ axis.

Step 5 ：Final Answer: The volume of the solid when the region bounded by the $y=\sqrt{4x}$ and $y=\sqrt{2-x}$ is rotated about the $x$ axis is $\boxed{{\displaystyle 6\pi }}$.