Chapter 6 | Applications of Integration
677
Figure 6.44 Calculating the lateral surface area of a frustum of a cone.
The cross-sections of the small cone and the large cone are similar triangles, so we see that r 2 r 1 = s − l s . Solving for s , we get
r 2 r 1
= s − l s
r 2 s = r 1 ( s − l ) r 2 s = r 1 s − r 1 l r 1 l = r 1 s − r 2 s r 1 l = ( r 1 − r 2 ) s r 1 l r 1 − r 2 = s .
Then the lateral surface area (SA) of the frustum is
S = (Lateral SA of large cone) −(Lateral SA of small cone) = πr 1 s − πr 2 ( s − l ) = πr 1 ⎛ ⎝ r 1 l r 1 − r 2 ⎞ ⎠ − πr 2 ⎛ ⎝ r 1 l r 1 − r 2 − l ⎞ ⎠
πr 1 2 l r 1 − r 2 2 l r 1 − r 2 πr 1 πr 1 2 l r 1 − r 2
πr 1 r 2 l r 1 − r 2 πr 1 r 2 l r 1 − r 2 πr 1 r 2 l r 1 − r 2
=
−
+ πr 2 l
πr 2 l ( r 1 − r 2 ) r 1 − r 2
=
−
+
πr 2 2 l r 1 − r 2
πr 1 r 2 l r 1 − r 2
=
−
+
−
π ⎛
2 2 ⎞
⎝ r 1 2 − r r 1 − r 2
⎠ l
π ( r 1 − r 2 )( r 1 + r 2 ) l r 1 − r 2
= = π ( r 1 + r 2 ) l . Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the x -axis. A representative band is shown in the following figure. =
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