Explanation Given information: Given the equation of hyperbola is y 2 − x 2 = 1 . Formula used: Volume of the solid using disk method is given by V = π ∫ a b y 2 d x . Calculation: From the below figure Consider a strip parallel to y axis then y = 1 + x 2 vary from x = − a to x = a . Then the volume of solid rotating about x axis for − a ≤ x ≤ a is V = π ∫ a b ( 1 + x 2 ) d x V = π [ x + x 3 3 ] − a <

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 6.3, Problem 62E

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To find the volume of the solid obtained by rotating the region between the branches of hyperbola y2−x2=1 about x axis for −a≤x≤a.

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Chapter 6.3, Problem 61E

Chapter 6.3, Problem 63E

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Chapter 6 Solutions

Calculus: Early Transcendentals

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Tocalculate: To find the volume of the solid obtained by rotating the region between the branches of hyperbola y 2 − x 2 = 1 about x axis for − a ≤ x ≤ a .

Solution Summary: The author calculates the volume of the solid obtained by rotating the region between the branches of hyperbola.

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Tocalculate:

To find the volume of the solid obtained by rotating the region between the branches of hyperbola y2−x2=1 about x axis for −a≤x≤a.

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Chapter 6 Solutions

Tocalculate: To find the volume of the solid obtained by rotating the region between the branches of hyperbola y 2 − x 2 = 1 about x axis for − a ≤ x ≤ a .

Solution Summary: The author calculates the volume of the solid obtained by rotating the region between the branches of hyperbola.

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Tocalculate: To find the volume of the solid obtained by rotating the region between the branches of hyperbola y2−x2=1 about x axis for −a≤x≤a.

Want to see the full answer?

Chapter 6 Solutions

Tocalculate:

To find the volume of the solid obtained by rotating the region between the branches of hyperbola y2−x2=1 about x axis for −a≤x≤a.