Explanation Given: (1) From the Exercise 10.2.53, the length of the ellipse x = a sin θ , y = b cos θ , where a > b > 0 , is L = 4 a ∫ 0 π 2 1 − e 2 sin 2 θ d θ where e = a 2 − b 2 a is the eccentricity of the ellipse. Here, the integrand is 1 − e 2 sin 2 θ . (2) From the Exercise 7.1.50, ∫ 0 π 2 sin 2 n θ d θ = 1 ⋅ 3 ⋅ 5 ⋯ ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 ⋯ 2 n π 2 . Result used: The Binomial series: If k is any number and | x | < 1 then, ( 1 + x ) k can be expressed as ( 1 + x ) k = ∑ n = 0 ∞ ( k n ) x n = 1 + k x + k ( k − 1 ) 2 ! x 2 + k ( k − 1 ) ( k − 2 ) 3 ! + ⋯ (1) Calculation: Substitute − x 2 for x and 1 2 for k in equation (1), ( 1 − x 2 ) 1 2 = ∑ n = 0 ∞ ( 1 2 n ) − x 2 n = 1 − 1 2 x 2 + 1 2 ( 1 2 − 1 ) 2 ! x 4 − 1 2 ( 1 2 − 1 ) ( 1 2 − 2 ) 3 ! x 6 + ⋯ = 1 − 1 2 x 2 + 1 2 ( − 1 2 ) 2 ! x 4 − 1 2 ( − 1 2 ) ( − 3 2 ) 3 ! x 6 + ⋯ = 1 − 1 2 x 2 − 1 2 2 1 2 ! x 4 − 1 2 3 1 ⋅ 3 3 ! x 6 + ⋯ That is, ( 1 − x 2 ) 1 2 = 1 − 1 2 x 2 − ∑ n = 2 ∞ 1 ⋅ 3 ⋅ 5 ⋯ ( 2 n − 3 ) 2 n n ! x 2 n Substitute e sin θ for x , ( 1 − e 2 sin 2 θ ) 1 2 = 1 + 1 2 ( e sin θ ) − ∑ n = 2 ∞ 1 ⋅ 3 ⋅ 5 ⋯ ( 2 n − 3 ) 2 n n ! ( e sin θ ) 2 n (2) Consider the length of the ellipse L = 4 a ∫ 0 π 2 1 − e 2 sin 2 θ d θ (3) Substitute equation (2) in equation (3), L = 4 a ∫ 0 π 2 [ 1 − 1 2 ( e sin θ ) 2 − ∑ n = 2 ∞ 1 ⋅ 3 ⋅ 5 ⋯ ( 2 n − 3 ) 2 n n ! ( e sin θ ) 2 n ] d θ = 4 a [ ∫ 0 π 2 d θ − ∫ 0 π 2 1 2 ( e sin θ ) 2 d θ − ∫ 0 π 2 ∑ n = 2 ∞<

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Author: Stewart

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Chapter 11.10, Problem 86E

To determine

To expand: The integrand as a binomial series and express the length of the ellipse L=4a∫0π21−e2sin2θdθ as a series in powers of the eccentricity up to the term in e6 .

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Chapter 11 Solutions

EBK SINGLE VARIABLE CALCULUS: EARLY TRA

Ch. 11.1 - (a) What is a sequence? (b) What does it mean to...Ch. 11.1 - (a) What is a convergent sequence? Give two...Ch. 11.1 - List the first five terms of the sequence. 3....Ch. 11.1 - List the first five terms of the sequence. 4....Ch. 11.1 - List the first five terms of the sequence. 5....Ch. 11.1 - List the first five terms of the sequence. 6....Ch. 11.1 - List the first five terms of the sequence. 7....Ch. 11.1 - List the first five terms of the sequence. 8....Ch. 11.1 - Prob. 9E Ch. 11.1 - List the first five terms of the sequence. 10. a1...

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To expand: The integrand as a binomial series and express the length of the ellipse L = 4 a ∫ 0 π 2 1 − e 2 sin 2 θ d θ as a series in powers of the eccentricity up to the term in e 6 .

Solution Summary: The author describes the integrand as a binomial series and expresses the length of an ellipse.

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To expand: The integrand as a binomial series and express the length of the ellipse L=4a∫0π21−e2sin2θdθ as a series in powers of the eccentricity up to the term in e6 .

Chapter 11 Solutions