(a) Explanation Given information: The curve function is y = ∫ 1 x t 2 − 1 d t . The interval is 1 ≤ x ≤ 5 . Revolution about x axis: If the function f ( x ) ≥ 0 is continuously differentiable on [ a , b ] the area of the surface generated by revolving the graph of y = f ( x ) about the x axis is S = ∫ a b 2 π y 1 + ( d y d x ) 2 d x Calculation: Show the curve Equation as follows: y = ∫ 1 x t 2 − 1 d t (1) Differentiate Equation (1) with respect to x . d y d x = x 2 − 1 Take square on both sides. ( d y d x ) 2 = ( x 2 − 1 ) 2 = x 2 − 1 Set up an integral for the area of the surface as shown below: S = ∫ a b 2 π y 1 + ( d y d x ) 2 (b) To determine Sketch the graph for the function y = ∫ 1 x t 2 − 1 d t . (c) To determine Calculate the surface area numerically by using integral evaluator.

14th Edition

ISBN: 9780135420683

Author: Hass

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Question

Chapter 6.4, Problem 8E

(a)

To determine

Form an integral for the surface area generated by revolving the curve about x axis.

(b)

To determine

Sketch the graph for the function y=∫1xt2−1dt.

(c)

To determine

Calculate the surface area numerically by using integral evaluator.

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Chapter 6.4, Problem 7E

Chapter 6.4, Problem 9E

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