Need full solution for (a) and (b) part only with
calculation analysis and detailed explanation with theorems as soon
as possible. I will give you thumbs up and a good
review.
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 radii r. R and height 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) is SA= (R+r) (R-1)2 + 12 (1) Let f : [a, b] → R be a continuously differentiable function, which means that f' is continuous on (1,6). S is differentiable on (a,b), and f 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 = 10,...,b=2n} of [0, b) and approximate the solid of revolution by conical frusta. Estimate the surface area SAS,P) of the solid rotated by f(x) on (a,b).
(b) Show that what you get from (a) is equivalent to SAU,P)=f(s) + (0-1)/1+ [!"()?Ax for some a € (101-16)
Need full solution for (a) and (b) part only with calculation analysis and detailed explanation with theorems as soon as
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