Let S be the region enclosed by the graphs of y = 2x and y = 2x2 for 0≤𝑥≤1. What is the volume of the solid generated when S is revolved about the line y = 3? ...3?

To find the volume of the solid generated when region S is revolved about the line y = 3, we can use the method of cylindrical shells. The formula for the volume of a solid of revolution using cylindrical shells is V = ∫2πrh dx, where r is the radius of the shell and h is the height of the shell.

Step 1: Identify the radius and height of the shells. The radius of the shell is the distance from the line y = 3 to the x-axis, which is |3 - y|. Since we are revolving around y = 3, the radius changes depending on whether we are above or below this line. For y = 2x (which is above the line y = 3 for 0 ≤ x ≤ 1), the radius is 3 - 2x. For y = 2x^2 (which is below the line y = 3 for 0 ≤ x ≤ 1), the radius is 3 - 2x^2. The height of the shell is the difference in y-values of the two functions, which is |2x - 2x^2|.

Step 2: Set up the integral. We need to integrate from x = 0 to x = 1. The volume is then given by V = ∫ from 0 to 1 [2π * (3 - 2x) * (2x - 2x^2) - 2π * (3 - 2x^2) * (2x - 2x^2)] dx.

Step 3: Evaluate the integral. This integral can be simplified and then evaluated using standard techniques of integration. The exact result will depend on the specific form of the integral after simplification.