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Re: Derivation for surface area of revolution
Posted:
May 25, 2012 11:41 AM


Thank you all for your replies, but I guess we're having a miscommunication.
First, I do not find the presentation of the derivation of the surface areaofrevolution formula in calculus books to be unclear or otherwise poorly presented. I just find them lacking in their common avoidance of addressing a very natural question that arises in the derivation.
Let me again try to explain why I find the responses I've read on this thread similarly lacking. Yes, to some it may seem "obvious" that you can't add up the surface area of infinitesimally thin cylinders (i.e. using dx instead of ds), and that curvature needs to be respected more explicitly. However, you could equally well make that "intuitive" argument in the derivation of the formula for volumes of revolution as well. Why does "dx" suffice (thence at first glance completely ignoring curvature of the curve revolved) in that case but not in the surface area formula? Obviously, it has something to do with the difference between calculating volumes and areas. In some sneakier way the curvature is accounted for in the volume calculation by the crosssectional areas being added up. But this is never clearly explicated in calculus books, nor have I seen here anything except standard "intuitive" arguments (and even that you won't find in today's namebrand calculus books). Perhaps you all don't find these arguments lacking, but I do. They're certainly not suitable for serious, inquisitive students. They're good having waving explanations, but that's all, in my opinion (note: I don't mean to be overly critical, as I don't have a better way to explain this either!)



