1. **State the problem:** We are given two curves: $x=2\sqrt{y}$ and $y=2\sqrt{x}$. We need to find the volume generated by revolving the area enclosed by these curves about the x-axis. 2. **Find the points of intersection:** Rewrite $x=2\sqrt{y}\implies \sqrt{y}=\frac{x}{2}\implies y=\frac{x^2}{4}$. Given $y=2\sqrt{x}$, square both sides to get $y^2=4x$. To find the intersection, substitute $y=\frac{x^2}{4}$ into $y=2\sqrt{x}$: $$\frac{x^2}{4} = 2\sqrt{x}$$ Multiply both sides by 4: $$x^2 = 8\sqrt{x}$$ Rewrite $\sqrt{x}$ as $x^{1/2}$: $$x^2 = 8x^{1/2}$$ Divide both sides by $x^{1/2}$ (assuming $x>0$): $$x^{3/2} = 8$$ Raise both sides to the power $\frac{2}{3}$: $$x = 8^{2/3} = (2^3)^{2/3} = 2^{2} = 4$$ Substitute back to get $y$: $$y = 2 \sqrt{4} = 2 \times 2 = 4$$ So the curves intersect at points $(0,0)$ and $(4,4)$. 3. **Set up the volume integral:** The volume generated by revolving around the x-axis is found by the washer method: $$V = \pi \int_0^4 \left[(\text{outer radius})^2 - (\text{inner radius})^2\right] dx$$ In terms of $x$, the outer radius is the y-value of the upper curve from the x-axis, and inner radius is the y-value of the lower curve. From the given curves: - Curve 1: $y=\frac{x^2}{4}$ - Curve 2: $y=2\sqrt{x}$ Between $x=0$ and $x=4$, $y=2\sqrt{x} \geq \frac{x^2}{4}$. So, $$V = \pi \int_0^{4} \left[(2\sqrt{x})^2 - \left(\frac{x^2}{4}\right)^2\right] dx = \pi \int_0^{4} \left[4x - \frac{x^4}{16}\right] dx$$ 4. **Evaluate the integral:** Calculate separately: $$\int_0^{4} 4x \, dx = 4 \times \frac{x^2}{2} \bigg|_0^4 = 4 \times \frac{16}{2} = 4 \times 8 = 32$$ $$\int_0^{4} \frac{x^4}{16} \, dx = \frac{1}{16} \times \frac{x^5}{5} \bigg|_0^4 = \frac{1}{16} \times \frac{1024}{5} = \frac{1024}{80} = \frac{128}{10} = 12.8$$ So, $$V = \pi (32 - 12.8) = \pi \times 19.2 = 19.2\pi$$ 5. **Final answer:** $$\boxed{19.2\pi}$$