To find the area of the surface generated when the given curve is revolved about the y-axis, we can use the formula for the surface area of revolution:
\[S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy\]
In this case, the curve is defined as \(x = 1 - 3y^3\) and the range of y is from 0 to 2.
First, we need to find \(\frac{dx}{dy}\). Taking the derivative of \(x\) with respect to \(y\), we get:
\[\frac{dx}{dy} = -9y^2\]
Next, we substitute the values into the formula:
\[S = \int_{0}^{2} 2\pi y \sqrt{1 + (-9y^2)^2} dy\]
Simplifying the expression inside the square root:
\[1 + (-9y^2)^2 = 1 + 81y^4\]
We can rewrite this as:
\[S = \int_{0}^{2} 2\pi y \sqrt{1 + 81y^4} dy\]
Finally, we can integrate this expression to find the area of the surface. However, as stated in the question, we are only asked to set up the integral and not evaluate it. Therefore, the integral setup is:
\[S = \int_{0}^{2} 2\pi y \sqrt{1 + 81y^4} dy\]