Find the area of the surface generated by revolving the curve about the indicated axis. \[ y=\frac{1}{6}\left(x^{2}+1\right), 0 \leq x \leq \sqrt{11}, y \text {-axis } \] Find the area of the surface $S$. $S=\square$ square units (Type an exact answer, using $\pi$ as needed.)

Step 1 ：The formula for the surface area of a curve \(y=f(x)\), \(a \leq x \leq b\), rotated about the y-axis is given by \(A = 2\pi \int_{a}^{b} x \sqrt{1+(f'(x))^2} dx\).

Step 2 ：First, we need to find the derivative of the function \(y=1/6(x^2+1)\). The derivative \(f'(x)\) is \(f'(x) = x/3\).

Step 3 ：Substitute \(f'(x)\) into the formula, we get \(A = 2\pi \int_{0}^{\sqrt{11}} x \sqrt{1+(x/3)^2} dx\).

Step 4 ：Simplify the integral, we get \(A = 2\pi \int_{0}^{\sqrt{11}} x \sqrt{1+x^2/9} dx\).

Step 5 ：To solve the integral, we can use the substitution method. Let \(u = 1+x^2/9\), then \(du = 2x/9 dx\) and \(dx = 9du / (2x)\).

Step 6 ：Substitute \(u\) and \(dx\) into the integral, we get \(A = 2\pi \int_{1}^{1+11/9} \sqrt{u} 9du / 2\).

Step 7 ：Solve the integral, we get \(A = 2\pi [2/3 u^{3/2} * 9/2]_{1}^{1+11/9}\).

Step 8 ：Calculate the definite integral, we get \(A = 2\pi [2/3 * (1+11/9)^{3/2} * 9/2 - 2/3 * 1^{3/2} * 9/2]\).

Step 9 ：Simplify the expression, we get \(A = 2\pi [2/3 * 1.732 * 9/2 - 2/3 * 1 * 9/2]\).

Step 10 ：Calculate the final result, we get \(A = 2\pi [2.31 - 1.5]\).

Step 11 ：Finally, we get \(A = 2\pi [0.81]\), so the area of the surface obtained by rotating the curve from \(x=0\) to \(x=\sqrt{11}\) about the \(y\)-axis is \(\boxed{1.62\pi}\) square units.