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Error Control for the Polar Area Formula Suppose that we wish to derive a formula for finding the area of the region 0 ≤ r ≤ f (θ) a≤θ≤b Call this area A. Suppose further that f (θ) ≤ M |f ′ (θ)| ≤ L for all a ≤ θ ≤ b. Divide the interval a ≤ θ ≤ b into n equal subintervals, each of length ∆θ = b−a . n Let θi∗ be the midpoint of the ith interval. On the ith interval, θ runs from θi∗ − 12 ∆θ to θi∗ + 12 ∆θ and the radius runs over all values of f (θ) with θi∗ − 12 ∆θ ≤ θ ≤ θi∗ + 12 ∆θ. Because |f ′ (θ)| ≤ L all of these values of f (θ) lie between ri = f (θi∗ ) − 21 L∆θ and Ri = f (θi∗ ) + 21 L∆θ. y θ=b y Ri θ=a ri x x So the area of the region 0 ≤ r ≤ f (θ), θi∗ − 12 ∆θ ≤ θ ≤ θi∗ + 21 ∆θ must lie between ∗ 2 2 2 1 1 1 and 12 ∆θ Ri2 = 12 ∆θ f (θi∗ ) + 21 L∆θ 2 ∆θ ri = 2 ∆θ f (θi ) − 2 L∆θ Observe that f (θi∗ ) ± 21 L∆θ implies that, since f (θ) ≤ M , Hence 2 = f (θi∗ )2 ± Lf (θi∗ )∆θ + 41 L2 ∆θ 2 2 f (θi∗ )2 − LM ∆θ + 14 L2 ∆θ 2 ≤ f (θi∗ ) ± 21 L∆θ ≤ f (θi∗ )2 + LM ∆θ + 41 L2 ∆θ 2 ∗ 2 1 2 f (θi ) ∆θ − 21 LM ∆θ 2 + 81 L2 ∆θ 3 ≤ area of sector #i ≤ 21 f (θi∗ )2 ∆θ + 12 LM ∆θ 2 + 81 L2 ∆θ 3 and the total area A obeys n h n h i i X X ∗ 2 2 3 ∗ 2 2 3 1 1 1 1 2 1 1 2 ≤ A ≤ f (θ ) ∆θ − LM ∆θ + L ∆θ f (θ ) ∆θ + LM ∆θ + L ∆θ i i 2 2 8 2 2 8 i=1 n X 1 f (θi∗ )2 ∆θ 2 i=1 c Joel Feldman. − 12 nLM ∆θ 2 + 81 nL2 ∆θ 3 ≤ A ≤ 2001. All rights reserved. i=1 n X 1 f (θi∗ )2 ∆θ 2 + 12 nLM ∆θ 2 + 18 nL2 ∆θ 3 i=1 1 Since ∆θ = 1 2 n X b−a n , f (θi∗ )2 ∆θ − 2 LM (b−a) 2 n 3 L2 (b−a) 8 n2 + ≤A≤ i=1 n X ∗ 2 1 2 f (θi ) ∆θ + 2 LM (b−a) 2 n + 3 L2 (b−a) 8 n2 i=1 Now take the limit as n → ∞. Since lim n→∞ n hX 1 f (θi∗ )2 ∆θ 2 ± 2 LM (b−a) 2 n + 3 L2 (b−a) 8 n2 i=1 = = Z b 1 2 b 1 2 Z i 2 LM (b−a) n n→∞ 2 f (θ)2 dθ ± lim a 3 L2 (b−a) 2 n n→∞ 8 + lim f (θ)2 dθ a we have that A= 1 2 Z b f (θ)2 dθ a exactly. c Joel Feldman. 2001. All rights reserved. 2