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Chapter 6 | Applications of Integration
Figure 6.45 A representative band used for determining surface area.
Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have S = π ( r 1 + r 2 ) l = π ⎛ ⎝ f ( x i −1 )+ f ( x i ) ⎞ ⎠ Δ x 2 + ⎛ ⎝ Δ y i ⎞ ⎠ 2 = π ⎛ ⎝ f ( x i −1 )+ f ( x i ) ⎞ ⎠ Δ x 1+ ⎛ ⎝ Δ y i Δ x ⎞ ⎠ 2 . Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select x i * ∈ [ x i −1 , x i ] such that f ′( x i * ) = ⎛ ⎝ Δ y i ⎞ ⎠ /Δ x . This gives us S = π ⎛ ⎝ f ( x i −1 )+ f ( x i ) ⎞ ⎠ Δ x 1+ ⎛ ⎝ f ′( x i * ) ⎞ ⎠ 2 . Furthermore, since f ( x ) is continuous, by the Intermediate Value Theorem, there is a point x i * * ∈ [ x i −1 , x i ] such that f ( x i * * ) = (1/2) ⎡ ⎣ f ( x i −1 )+ f ( x i ) ⎤ ⎦ , so we get S =2 πf ( x i * * )Δ x 1+ ⎛ ⎝ f ′( x i * ) ⎞ ⎠ 2 . Then the approximate surface area of the whole surface of revolution is given by Surface Area ≈ ∑ i =1 n 2 πf ( x i * * )Δ x 1+ ⎛ ⎝ f ′( x i * ) ⎞ ⎠ 2 . This almost looks like a Riemann sum, except we have functions evaluated at two different points, x i * and x i * * , over the interval [ x i −1 , x i ]. Although we do not examine the details here, it turns out that because f ( x ) is smooth, if we let n →∞, the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both x i * and x i * * are in the interval [ x i −1 , x i ], so it makes sense that as n →∞, both x i * and x i * * approach x . Those of you who are interested in the details should consult an advanced calculus text. Taking the limit as n →∞, we get Surface Area = lim n →∞ ∑ i =1 n 2 πf ( x i * * )Δ x 1+ ⎛ ⎝ f ′( x i * ) ⎞ ⎠ 2 = ∫ a b ⎛ ⎝ 2 πf ( x ) 1+ ⎛ ⎝ f ′( x ) ⎞ ⎠ 2 ⎞ ⎠ dx . As with arc length, we can conduct a similar development for functions of y to get a formula for the surface area of surfaces of revolution about the y -axis. These findings are summarized in the following theorem.
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