Surface area of a solid of revolution

[newageanubis]

Homework Statement

Having recently learned the disk/shell/washer method for finding the volume of a solid of revolution, I'm trying to apply similar methods to derive the formula for the surface area of a cone (and hopefully after that, that of a sphere).
The region that is revolved around to form the cone is that under f(x) = (r/h)x from 0 to h, where r is the radius of the base and h is the height of the cone (both are constants).

Homework Equations

Since V = πr^2h for a cylinder, the volume of the cone is ∫ π[f(x)]^2 dx from 0 to h. When I evaluate that, I get V = πr^2h/3,which is correct.

The Attempt at a Solution

I reasoned that since A = 2πrh for the curved surface of a cylinder, evaluating ∫ 2πf(x) dx from 0 to h should result in an expression for the area of the curved surface of the cone (everything but the base). But instead of getting πrs (s being √(r^2 + h^2), I think it's called the lateral height), I get πrh as the area of the curved surface.

Is my method wrong, or am I just integrating wrong?

 

[tiny-tim]

hi newageanubis!

I reasoned that since A = 2πrh for the curved surface of a cylinder …

yes, but the surface of a cone is sloping, so it's a lot more than that, isn't it?

 

[newageanubis]

Oh. Now I feel dumb :(.

Can this "problem" be solved with single-variable integration, though?

 

[tiny-tim]

you'd be surprised how often people ask the same thing on this forum!

Can this "problem" be solved with single-variable integration, though?

not following you, this is a single ∫, isn't it?

 

[HallsofIvy]

If a curve is given by y= f(x), then a "differential of arc length" is given by ds= \sqrt{(dx)^2+ (dy)^2}= \sqrt{1+ (dy/dx)^2}dx.

And it is the arclength, rotated around an axis, that will give you a surface area.

In the case of the cone, say y= ax, rotated around the y- axis, the circumference of a small section would be 2\pi x so that it area would be 2\pi x\sqrt{1+ (dy/dx)^2}dr. Of course, in this example, dy/dx= a so that would be 2\pi x\sqrt{1+ a^2}dx. The area from the coordinate plane up to y= aR, so that x= R, would be \pi \int_0^R x\sqrt{1+ a^2}dx= \pi R^2\sqrt{1+ a^2}/2.