Explanation Given information: The solid bounded by the region x 2 − y 2 = a 2 and x = a + h about the y axis is ( a > 0 , h > 0 ) . Find the intersection points of the curves as shown below: x 2 − y 2 = a 2 y 2 = x 2 − a 2 y = ± x 2 − a 2 Consider the condition x = a ( a > 0 ) . The region enclosed by the curves y = x 2 − a 2 and y = − x 2 − a 2 shown in Figure 1. Refer to Figure (1). Consider that the solid is a washer which is obtained by rotating region, about the line y axis. Calculate the volume rotate about y –axis. V = 2 π ∫ a a + h x y 2 d x − 2 π ∫ a a + h x y 1 d x (1) Here, a is lower limit for the integral and a + h is upper limit for the integral Substitute x 2 − a 2 for y 2 , and − x 2 − a 2 for y 2 in Equation (1) V = 2 π ∫ a a + h x x 2 − a 2 d x − 2 π ∫ a a + h x ( − x 2 − a 2 ) d x Modify Equation (1), V = 2 π ∫ a a + h x x 2 − a 2 d x + 2 π ∫ a a + h x ( x 2 − a 2 ) d x = 4 π ∫ a a + h x x 2 − a 2 d x Consider u = x 2 − a 2 (2) Differentiate Equation (2) with respect to x

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ISBN: 9781337613927

Author: Stewart

Publisher: CENGAGE L

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Chapter 6, Problem 11E

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Chapter 6 Solutions

Calculus, Early Transcendentals

Ch. 6.1 - (a) Set up an integral for the area of the shaded...Ch. 6.1 - (a) Set up an integral for the area of the shaded...Ch. 6.1 - Find the area of the shaded region.Ch. 6.1 - Find the area of the shaded region.Ch. 6.1 - Find the area of the shaded region. 5.Ch. 6.1 - Find the area of the shaded region. 6.Ch. 6.1 - Set up, but do not evaluate, an integral...Ch. 6.1 - Prob. 8E Ch. 6.1 - Set up, but do not evaluate, an integral...Ch. 6.1 - Set up, but do not evaluate, an integral...

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The volume of the solid obtained by rotating region.

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