1. Find the area of the surface generated by revolving y = (sqrt)(2x-x^2), x = [0,1] a. About the x axis b. About the y-axis This problem is from the section about surface area revolving around an axis. I actually did the problem but im not sure on my answer. For around the x axis I got an answer of 2pi. The integral that I set up was integrating 2pi as a constant and then dx from 0 to 1. For the b part I got the integral of 2pi as a constant and then integrating x / (sqrt)(2x-x^2) Please see if this is correct, if not please explain to me how it should be done. Thank you!