Explanation Given: A curve is given as c ( t ) = ( ( t + 1 ) 3 , ln t ) , 1 ≤ t ≤ 5 Formula used: The surface area for the surface revolving by the curve [ c ( t ) = ( x ( t ) , y ( t ) ) ] about x-axis for a ≤ t ≤ b is given by S = 2 π ∫ b a y ( t ) x ′ ( t ) 2 + y ′ ( t ) 2 d t Simpson’s formula for N subintervals is given by S N = ∫ a b f ( x ) d x = Δ x 3 ( f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + ....... + 2 f ( x N − 2 ) + 4 f ( x N − 1 ) + f ( x N ) ) , w h e r e , Δ x = b − a N a n d x i = x i − 1 + Δ x , 1 ≤ i ≤ N { x 0 = a a n d x N = b } Calculation: Here, the curve is given by x ( t ) = ( t + 1 ) 3 ; y ( t ) = ln t Then, by differentiating we get x ′ ( t ) = 3 ( t + 1 ) 2 , y ′ ( t ) = 1 t Now, surface area can be calculated as S = 2 π ∫ b a y ( t ) x ′ ( t ) 2 + y ′ ( t ) 2 d t = 2 π ∫ 1 5 ln t ( 9 ( t + 1 ) 4 + 1 t 2 ) d t For interval [ 1 , 5 ] and N = 50 , Δ x = 5 − 1 50 = 4 50 = 2 25 So, the points can be calculated as x 0 = 1 , x 1 = 1 + 2 25 = 27 25 , x 2 = 27 25 + 2 25 = 29 25 , x 3 = 29 25 + 2 25 = 31 25 , x 4 = 31 25 + 2 25 = 33 25 , and so on. Here, f ( t ) = ln t ( 9 ( t + 1 ) 4 + 1 t 2 ) Thus, the value of S 50 can be calculated as shown below

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Chapter 11.2, Problem 40E

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To compute:

The surface area of the given curve using Simpson’s Rule.

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Chapter 11.2, Problem 39E

Chapter 11.2, Problem 41E

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Chapter 11 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 6PQ Ch. 11.1 - Prob. 7PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E

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The surface area of the given curve using Simpson’s Rule.

Solution Summary: The author calculates the surface area of a given curve using Simpson's Rule, which is 1612.09484 square unit.

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To compute: The surface area of the given curve using Simpson’s Rule.

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Chapter 11 Solutions

To compute:

The surface area of the given curve using Simpson’s Rule.