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Derivation for surface area of revolution
Posted:
May 22, 2012 5:23 PM


Deriving the integral formula for the surface area of a curve revolved around an axis often raises the following question (and suppose y=f(x) is revolved around the xaxis):
"Why do we multiply 2*pi*f(x) by the arc length differential ds instead of the differential dx?"
This is a perfectly reasonable question since for volumes of revolution one can multiply a cross section area A(x) by dx, and then integrate A(x)*dx to get the volume. No need for worrying about the curvature that ds captures. The plain old dx differential does just fine.
Almost every modern calculus book out there bypasses this issue, which I find rather shameless. I've had a look at Anton, Thomas, Stewart, and a host of other standard texts and they say nothing about it. Any explanations I've seen amount to "well, if you use dx instead of ds then you get the wrong answer; calculating surface area is different than volume."
Who could dispute that standard answer? Using dx instead of ds just doesn't recover the proper formulas for surface areas of simple surfaces. Fine. But doesn't anyone find that "proof is in the pudding" explanation dissatisfying? Isn't there a better explanation than this? Why does adding up (integrating) A(x)*dx work for volumes but adding up 2*pi*f(x)*dx doesn't work for surface areas? Is there a more intuitive explanation?



