1. **State the problem:** We want to find the volume $V$ by integrating the volume element $dV = \pi r^2 dh$ where $r = y = \frac{2}{x}$ and $dh = dx$. 2. **Write the formula:** The volume element is given by $$dV = \pi r^2 dh = \pi \left(\frac{2}{x}\right)^2 dx = \pi \frac{4}{x^2} dx = 4\pi \frac{1}{x^2} dx.$$ 3. **Set up the integral:** $$V = \int dV = \int 4\pi \frac{1}{x^2} dx.$$ 4. **Integrate:** Recall that $$\int x^{-2} dx = \int \frac{1}{x^2} dx = -\frac{1}{x} + C.$$ So, $$V = 4\pi \int x^{-2} dx = 4\pi \left(-\frac{1}{x} + C\right) = -\frac{4\pi}{x} + C'.$$ 5. **Interpretation:** The volume function is $$V = -\frac{4\pi}{x} + C',$$ where $C'$ is the constant of integration determined by boundary conditions. **Final answer:** $$V = -\frac{4\pi}{x} + C'$$