Find the area of the surface generated by revolving $x=t+2 \sqrt{5}, y=\frac{t^{2}}{2}+2 \sqrt{5} t+2,-2 \sqrt{5} s t \leq 2 \sqrt{5}$ about the $y$-axis. The surface area obtained by revolving the given curve around the $y$-axis is (Type an exact answer in terms of $\pi$.)

Step 1 ：Given the curve defined by the parametric equations \(x = t + 2\sqrt{5}\) and \(y = \frac{t^2}{2} + 2\sqrt{5}t + 2\), we want to find the area of the surface generated by revolving this curve about the y-axis.

Step 2 ：The formula for the surface area of a solid of revolution about the y-axis is given by \(A = 2\pi \int_{a}^{b} x \sqrt{1 + (\frac{dx}{dy})^2} dy\), where \(x = f(y)\) is the equation of the curve, and \(a\) and \(b\) are the limits of \(y\).

Step 3 ：We first need to express \(x\) in terms of \(y\) and find \(\frac{dx}{dy}\). To do this, we solve the equation \(y = \frac{t^2}{2} + 2\sqrt{5}t + 2\) for \(t\) and substitute it into \(x = t + 2\sqrt{5}\) to get \(x\) in terms of \(y\).

Step 4 ：Solving for \(t\) gives us \(t = -\sqrt{2y + 16} - 2\sqrt{5}\) and \(t = \sqrt{2y + 16} - 2\sqrt{5}\). Substituting the first solution into \(x = t + 2\sqrt{5}\) gives us \(x = -\sqrt{2y + 16}\).

Step 5 ：We then find \(\frac{dx}{dy}\) by differentiating \(x\) with respect to \(y\), which gives us \(\frac{dx}{dy} = -\frac{1}{\sqrt{2y + 16}}\).

Step 6 ：Substituting \(x\) and \(\frac{dx}{dy}\) into the formula for the surface area and evaluating the integral from \(-2\sqrt{5}\) to \(2\sqrt{5}\) gives us the surface area \(A\).

Step 7 ：Finally, we find that the area of the surface generated by revolving the given curve about the y-axis is \(\boxed{2\pi\left(-2\left(2\sqrt{5} + 8\right)\sqrt{4\sqrt{5} + 17}/3 - \sqrt{4\sqrt{5} + 17}/3 + \sqrt{17 - 4\sqrt{5}}/3 + 2\left(8 - 2\sqrt{5}\right)\sqrt{17 - 4\sqrt{5}}/3\right)}\).