Hello. I am a calculus student currently studying integrals. I am able to logically break down most problems, but I came to a breaking point today. I am trying to understand how to revolve the area between two trigonometric functions about a vertical line. The conditions I am trying to satisfy is when these functions form an area that is apparently symmetric and end on identical y coordinates. To simplify my question, I'll create an example. Revolve the area enclosed by f(x) and g(x) about the line x= -1 on the interval [0, [tex]\pi[/tex]] f(x) = sin(x) g(x) = -sin(x) I am having incredible difficulty merely setting up the integral. Clearly, I'll need something like this: V = [tex]\pi[/tex] [tex]\int[/tex] ( something + 1)^2 - ( something + 1)^2 dy I understand the concept of selecting f(y) and g(y) and subtracting the axis' x-coordinate, but I do not understand how to choose which one is an outer radius and which one is an inner radius. Additionally, because the points that they intersect at are identical, I am not sure what bounds to use without getting a bad answer (v=0). I appreciate your time and hope that you will forgive my English.