Can anyone help me with finding the surface area of a solid of revolution of f(x) about the x axis for the interval [1,6]. It's supposed to be able to be done without needing calculus but I am having trouble figuring it out.
$f(x) = \begin{cases} 1 & 1 \leq x< 2\\ 1/2 & 2 \leq x< 3\\ . & .\\ . & .\\ 1/n & n\leq x< n+1\\ \end{cases}$
Any help is appreciated. Thanks
The solid consists of cylinders, where the radius of a particular cylinder is the value of $f$ at that part. The wanted surface area is the "outer parts" of the cylinders. Does this help?