- #1

- 1,179

- 5

The Volume of a revolved function can be given by the integral of pi*f(x)^2*dx. For the arc length of a graph, a different integral is available. I understood the proof of these two and their integration is understandable. From such, I was actually expecting that the surface area of revolution would be the integral of 2*pi*f(x)*dx, in which it would be a summation of infinitesmally thin hoops over an interval deltax. This actually seems logical to me...but why doesn't it work. Apparently, the surface area of revolution is given by the integral of 2*pi*f(x)*dL, in which dL is infinitesmally small arc length and can be replaced by a function in respect to dx...which makes things lengthier. Why doesn't integrating with just dx work???...it seems as though it would be the same concept of volume, just with the surface area instead.

Last edited: