Find the area of the surface obtained by rotating the curve
\[
y=1+6 x^{2}
\]
from $x=0$ to $x=4$ about the $y$-axis.

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 $y=1+6x^2$. The derivative $f'(x)$ is $f'(x) = 12x$.

Step 3 ：Substitute $f'(x)$ into the formula, we get $A = 2\pi \int_{0}^{4} x \sqrt{1+(12x)^2} dx$.

Step 4 ：Simplify the integral, we get $A = 2\pi \int_{0}^{4} x \sqrt{1+144x^2} dx$.

Step 5 ：To solve the integral, we can use the substitution method. Let $u = 1+144x^2$, then $du = 288x dx$ and $dx = du / (288x)$.

Step 6 ：Substitute $u$ and $dx$ into the integral, we get $A = 2\pi \int_{1}^{577} \sqrt{u} du / 288$.

Step 7 ：Solve the integral, we get $A = 2\pi [2/3 u^{3/2} / 288]_{1}^{577}$.

Step 8 ：Calculate the definite integral, we get $A = 2\pi [2/3 * 577^{3/2} / 288 - 2/3 * 1^{3/2} / 288]$.

Step 9 ：Simplify the expression, we get $A = 2\pi [2/3 * 138.56 - 2/3 * 1 / 288]$.

Step 10 ：Calculate the final result, we get $A = 2\pi [92.37 - 0.0023]$.

Step 11 ：Finally, we get $A = 2\pi [92.3677]$, so the area of the surface obtained by rotating the curve from $x=0$ to $x=4$ about the $y$-axis is $\boxed{581.68 \pi}$.