The key lies in the derivation of transverse area. The simplest method is to use the idea of limit to cross the frustum into countless small frustums, so that each frustum can be approximately regarded as a cylinder. Then, by using calculus, we can find the S-side = ∫ (0 to L) 2π dz = π (r1+R2) L.

Where am I?

Is the length of the frustum bus, r 1, r2 is the radius of the upper and lower circles, so S=S side +S upper +S lower = π (r1+R2) l.

+πr 12+πR22 =π(r ' 2+R2+r ' l+rl)

Of course, the formula of rotating surface area is used. S=2π∫ydx。

Where y=(r2-r 1)x/L+r 1.

S-edges can also be solved, but they are all advanced mathematics, and high school mathematics does not require the derivation of truncated cone surface area formula.

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