Please can someone explain to me why we use $dx$ in a volume of revolution i.e. $$\pi \int{f(x)^2 dx}$$ but $ds$ (an elementary bit of arc) in a surface of revolution i.e. $$2\pi \int{f(x)ds}$$ does this not mean you are under or over estimating the volume in the volume of revolution
2026-03-29 09:11:31.1774775491
Volumes and surfaces of revolution?
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as $dx \to 0$ the over/under-estimation of volume you note dwindles to zero. however this does not apply to the computation of surface area. you can see the same principle more clearly if you compare the usual procedure for computing the area under a curve with the requirements of computing an arc length for the same function.