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Derivation for surface area of revolution
Posted:
May 22, 2012 5:23 PM
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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 x-axis):
"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?
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