This isn't homework. It's maths curiosity. I have a hair curler set. The curlers are spool shape, 3 inches long, 1/2 inch diameter ends, 3/8 inch at the middle. The surface is a surface of revolution. The curve generating the surface is a circular arc, concave on the curler. Assume a smooth surface and ignore the end area. I have a picture of the curler at the bottom of my main webpage. http://www.geocities.com/fran3406/Cee_Ann.html Here's how I did it, but I want to know how to do it using Calculus. I don't know Calculus except for some of the really basics but I should be able to understand a solution for this problem. Okay, If I roll the curler out one circumference and imagine the net that would roll back up to make the surface of the curler, I get a rectangular area which is 3 by 0.5pi minus a parabola having base 3 and height 0.5pi - 0.375pi. Imagine all the circumferences from end to end laid out next to each other to form the total area which should be a rectangle with two parabolic notches along each of the 3 inch ends. You could have one parabola of the calculated height or two parabolas half that height ... total area in each case is the same. The height of the single parabola is 0.392699081, so its area is 2/3 of 3 times 0.392699081 which equals 0.785398161 which is subtracted from 3 times 0.5pi to get a surface area of 3.926990819 square inches. If this is correct, it simplifies to (3/4pi)/3.