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4. In this problem, we would like to find the surface area of a solid of revolution (a) The solid showed in the picture is called the conical frustum with radil r. R and height h. h R Suppose we know that the surface area of this conical frustum(only the curve surface area which means the top and bottom areas are not included See it in 3D space) SA = *(R+r) (R-1) + (1) Let f : [a, b] → R be a continuously differentiable function, which means that is continuous on [a,b], is differentiable on (a,b), and is continuous on (a,b). Consider the solid of revolution ob- tained by rotating the graph off about the z-axis. Choose a partition P = {a = 0, of [a, b] and approximate the solid of revolution by conical frusta. Estimate the surface area SA(S.P) of the solid rotated by f(x) on (a,b).

(b) Show that what you get from (s) is equivalent to TI SAS,P) = f(x) + f(1-1)/1+ (') Afor some 4 € (1-1)