Find the volume when the region bounded by the $y=\sqrt{4 x}$ 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 \(y=\sqrt{4 x}\) 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 \(y=\sqrt{4 x}\) and \(y=\sqrt{2-x}\) is rotated about the \(x\) axis is \(\boxed{6\pi}\).