1. Problem statement: Find the volume of the solid obtained by revolving the region bounded by $x=2\sqrt{y}$ and $y=2\sqrt{x}$ about the x-axis. 2. Convert the equation $x=2\sqrt{y}$ to a function of $x$ by solving for $y$, which gives $y=\frac{x^2}{4}$. 3. Identify the two curves as $y_{1}=\frac{x^2}{4}$ and $y_{2}=2\sqrt{x}$, and note we will revolve the region between them about the x-axis. 4. Find intersection points by solving $\frac{x^2}{4}=2\sqrt{x}$. 5. Let $u=\sqrt{x}$ so that $x=u^2$, and substitute to obtain $\frac{u^4}{4}=2u$. 6. Multiply both sides by 4 to get $u^4=8u$, which factors as $u(u^3-8)=0$. 7. Solve to get $u=0$ or $u^3=8$, hence $u=0$ or $u=2$, which gives $x=0$ or $x=4$ and corresponding $y$ values $0$ and $4$. 8. Determine the outer radius $R(x)$ and inner radius $r(x)$ for washers when revolving about the x-axis: $R(x)=2\sqrt{x}$ and $r(x)=\frac{x^2}{4}$. 9. Write the volume using the washer method: $$V=\pi\int_{0}^{4}\left[(2\sqrt{x})^2-\left(\frac{x^2}{4}\right)^2\right]dx$$. 10. Simplify the integrand by computing the squares: $(2\sqrt{x})^2=4x$ and $\left(\frac{x^2}{4}\right)^2=\frac{x^4}{16}$, so $$V=\pi\int_{0}^{4}\left(4x-\frac{x^4}{16}\right)dx$$. 11. Find an antiderivative: an antiderivative of $4x-\frac{x^4}{16}$ is $2x^2-\frac{x^5}{80}$. 12. Evaluate the definite integral from 0 to 4: $$\left[2x^2-\frac{x^5}{80}\right]_{0}^{4}=2(4^2)-\frac{4^5}{80}-0=32-\frac{1024}{80}$$. 13. Simplify $\frac{1024}{80}=\frac{64}{5}$, so the integral evaluates to $32-\frac{64}{5}=\frac{96}{5}$. 14. Multiply by $\pi$ to obtain the volume $V=\frac{96\pi}{5}$. 15. Final answer: $V=\frac{96\pi}{5}$.