- #1
mathFun
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So if we are looking to find the surface area of a solid of revolution formed by rotating a curve about a line, we can use the following:
S = ∫2(pi)yds (if rotating about the x-axis)
FOr example, say our curve is y=x2 and we want to find the surface area of the solid of revolution that's found when we rotate y about the x axis, on the interval from 0 to 1.
Intuitively, this makes sense for me except for one thing. If I think of this problem the same way I do for Reimann integrals, I can imagine slicing the x-axis up into tiny pieces, then for each of those intervals of width Δx I would take the circumference of the circle with radius f(x) and multiply this by Δx. This would give me the surface area of cylinder of radius f(x) and height Δx. So now if I take Δx infinitesimally small, it seems like I would end up kind of adding together these circumferences, and end up with the surface area.
Why is it though that we have to use ds (arc length parameter) then instead of using dx? Why doesn't it work to do it like I mentioned, where you cut up the x axis?
S = ∫2(pi)yds (if rotating about the x-axis)
FOr example, say our curve is y=x2 and we want to find the surface area of the solid of revolution that's found when we rotate y about the x axis, on the interval from 0 to 1.
Intuitively, this makes sense for me except for one thing. If I think of this problem the same way I do for Reimann integrals, I can imagine slicing the x-axis up into tiny pieces, then for each of those intervals of width Δx I would take the circumference of the circle with radius f(x) and multiply this by Δx. This would give me the surface area of cylinder of radius f(x) and height Δx. So now if I take Δx infinitesimally small, it seems like I would end up kind of adding together these circumferences, and end up with the surface area.
Why is it though that we have to use ds (arc length parameter) then instead of using dx? Why doesn't it work to do it like I mentioned, where you cut up the x axis?