Özet: The problem asks to calculate the area of the surface generated by revolving the curve about the x-axis. The curve is given by the parametric equations: $x(t) = t^2$ $y(t) = 2t$ for $0 \le t \le \sqrt{3}$. The formula for the area of a surface of revolution about the x-axis for a parametrically defined curve is: $S = \int_{t_1}^{t_2} 2\pi y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$ First, we need to find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $\frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t$ $\frac{dy}{dt} = \frac{d}{dt}(2t) = 2$ Next, we calculate the term inside the square root: $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (2t)^2 + (2)^2 = 4t^2 + 4$ Now, take the square root: $\sqrt{4t^2 + 4} = \sqrt{4(t^2 + 1)} = 2\sqrt{t^2 + 1}$ Substitute $y(t)$ and the square root term into the surface area formula: $S = \int_{0}^{\sqrt{3}} 2\pi (2t) (2\sqrt{t^2 + 1}) dt$ $S = \int_{0}^{\sqrt{3}} 8\pi t \sqrt{t^2 + 1} dt$ To evaluate this integral, we can use a substitution. Let $u = t^2 + 1$. Then, $du = 2t \, dt$. We also need to change the limits of integration: When $t = 0$, $u = 0^2 + 1 = 1$. When $t = \sqrt{3}$, $u = (\sqrt{3})^2 + 1 = 3 + 1 = 4$. Now, substitute these into the integral: $S = \int_{1}^{4} 8\pi \sqrt{u} \left(\frac{1}{2} du\right)$ $S = 4\pi \int_{1}^{4} u^{1/2} du$ Integrate $u^{1/2}$: $\int u^{1/2} du = \frac{u^{1/2 + 1}}{1/2 + 1} = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$ Now, evaluate the definite integral: $S = 4\pi \left[\frac{2}{3} u^{3/2}\right]_{1}^{4}$ $S = 4\pi \left(\frac{2}{3} (4)^{3/2} - \frac{2}{3} (1)^{3/2}\right)$ $S = 4\pi \left(\frac{2}{3} (\sqrt{4})^3 - \frac{2}{3} (1)^3\right)$ $S = 4\pi \left(\frac{2}{3} (2)^3 - \frac{2}{3}\right)$ $S = 4\pi \left(\frac{2}{3} (8) - \frac{2}{3}\right)$ $S = 4\pi \left(\frac{16}{3} - \frac{2}{3}\right)$ $S = 4\pi \left(\frac{14}{3}\right)$ $S = \frac{56\pi}{3}$ The final answer is $\boxed{\frac{56\pi}{3}}$.