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

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

Step 2 ：First, we need to find the derivative of the function $y=1+6x^2$. The derivative $f^{\prime }(x)$ is $f^{\prime }(x)=12x$.

Step 3 ：Substitute $f^{\prime }(x)$ 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 $du=288xdx$ 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 $A=2\pi [2/3u^{3/2}/288{]}_1^{577}$.

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

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

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 $\boxed{{\displaystyle 581.68\pi }}$.