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f(x) is a continuous function of x, whose domain is [a, b]. Revolve the graph around the x axis. In doing so you will create a solid.

Apparently, to find the volume of this solid, partition the solid into n cylinders along the x axis from a to b, each partition of the x axis containing some x*. A cylinder's radius is f(x*), so its volume is pi(f(x*)^2). Add the cylinders' volumes together. Take the limit as n approaches infinity; that is, take the integral of pi(f(x)^2)dx from x=a to x=b.

So that's volume. To find the area of the surface of revolution, instead of using cylinders, partition the solid into n

My question: Why the inconsistency? Why can't we take the area of the surface of revolution just by taking cylinders, like we did with volume (except of course taking the surface area of the cylinders, not the volume), and not have to worry about the arc length (that is, not have to worry about the fact that, for a given partition, one circular side will be different from the other, since you'd think it wouldn't matter when the partition becomes infinitely thin)? And if we're not allowed to do that, why are we allowed to use cylinders instead of frustums and ignore the arc length when taking volume?

In my calculus II class we're using Stewart's Calculus as our textbook. The book doesn't really explain the inconsistency. I also own Spivak's Calculus, which doesn't seem to explain it either (although Spivak at least admits there's some "fudging" in what he's doing in that chapter, although I'm not sure if this is what he was referring to or not).

I might not have explained my question well in the post. If you're confused about what I'm asking, I can try clearing it up.

Apparently, to find the volume of this solid, partition the solid into n cylinders along the x axis from a to b, each partition of the x axis containing some x*. A cylinder's radius is f(x*), so its volume is pi(f(x*)^2). Add the cylinders' volumes together. Take the limit as n approaches infinity; that is, take the integral of pi(f(x)^2)dx from x=a to x=b.

So that's volume. To find the area of the surface of revolution, instead of using cylinders, partition the solid into n

*frustums of cones*along the x axis from a to b, each frustum having two different circular sides, one with radius f(xi-1) and the other with radius f(xi). The surface area of a frustum is 2pi times the average of the radii times the arc length of the curve cut by the two circular sides of the frustum, this arc length being sqrt[1+(dy/dx)^2]. Add the frustums' surface areas together. Take the limit as n approaches infinity; that is, take the integral of 2pif(x)sqrt[1+(dy/dx)^2]dx from x=a to x=b.My question: Why the inconsistency? Why can't we take the area of the surface of revolution just by taking cylinders, like we did with volume (except of course taking the surface area of the cylinders, not the volume), and not have to worry about the arc length (that is, not have to worry about the fact that, for a given partition, one circular side will be different from the other, since you'd think it wouldn't matter when the partition becomes infinitely thin)? And if we're not allowed to do that, why are we allowed to use cylinders instead of frustums and ignore the arc length when taking volume?

In my calculus II class we're using Stewart's Calculus as our textbook. The book doesn't really explain the inconsistency. I also own Spivak's Calculus, which doesn't seem to explain it either (although Spivak at least admits there's some "fudging" in what he's doing in that chapter, although I'm not sure if this is what he was referring to or not).

I might not have explained my question well in the post. If you're confused about what I'm asking, I can try clearing it up.

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