Explanation Given: The functions are x = y 2 − 3 , x = 2 y , and axis y = 4 . Formula used: Volume V = ∫ c d A ( y ) d y Where c ≤ y ≤ d and A = 2 π r h . Calculation: The functions are x = y 2 − 3 , x = 2 y , and axis y = 4 . The graphs of x = y 2 − 3 and x = 2 y intersect where y 2 − 3 = 2 y y 2 − 3 − 2 y = 0 y 2 − 2 y − 3 = 0 Find the factors of the above equation. y 2 − 3 y + y − 3 = 0 y ( y − 3 ) + 1 ( y − 3 ) = 0 ( y − 3 ) ( y + 1 ) = 0 That is. ( y − 3 ) = 0 or ( y + 1 ) = 0 y = 3 or y = − 1 The graph of the function is: Thus, the enclosed area lies over the interval [ − 1 , 3 ] . Choose the slice of region bounded by the graphs x = y 2 − 3 and x = 2 y horizontally over the interval [ − 1 , 3 ] . This horizontal slice is parallel to the axis of rotation. Now, use the shell method to calculate the volume of the solid. For each y ∈ [ − 1 , 3 ] , the shell with radius 4 − y and height 2 y − ( y 2 − 3 ) = 3 + 2 y − y 2 ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 6, Problem 31CRE

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Find the volume of the solid obtained by rotating the region enclosed by the curves about the given axis.

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Chapter 6, Problem 30CRE

Chapter 6, Problem 32CRE

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