Final answer:
The student's question is about the setup of an integral for calculating the surface area of a cylindrical surface created by revolving the curve y = 2 about the x-axis from y = 0 to y = 1. The appropriate integral formula is 2\(\pi\)\(\int_{0}^{1} 2 dy\), which represents the circumference of the base circle times the height.
Step-by-step explanation:
The student is asking for assistance in setting up an integral to find the surface area of a solid of revolution. Given the curve y = 2, which is a horizontal line, and revolving this curve about the x-axis from y = 0 to y = 1, we are generating a cylindrical surface. The formula for the surface area of a shape generated by revolving a function f(y) about the x-axis is generally written as:
\( SA = 2\pi \int_{y_0}^{y_1} f(y) \sqrt{1 + \left( \frac{dx}{dy} \right)^2} dy \)
In this case, since y = 2 is a constant function, its derivative with respect to y is zero, simplifying our integral to:
\( SA = 2\pi \int_{0}^{1} 2 dy \)
We do not evaluate this integral as per the instruction in the question, but the set up implies the multiplication of the circumference of the circle (which is 2\(\pi\) times the radius, here radius is 2) by the height of the cylinder (1 unit, from y = 0 to y= 1).