Explanation Given : The region bounded by z = 1 − y 2 and z = y 2 − 1 for 0 ≤ x ≤ 2 . Calculation : Step 1: The volume of the region is the double integral of f ( y , z ) = 2 over the domain D in the yz-plane between the curves z = 1 − y 2 and z = y 2 − 1 . This is a vertically simple region described by the inequalities: − 1 ≤ y ≤ 1 , y 2 − 1 ≤ z ≤ 1 − y 2 Step 2: We compute the double integral as the following iterated integral: V = ∫ ∫ D 2 d A = ∫ − 1 1 ∫ y 2 − 1 1 − y 2 2 d z d y = ∫

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 16.2, Problem 46E

To determine

The volume of the region bounded by $z = 1 - y^{2}$ and $z = y^{2} - 1$ for $0 \leq x \leq 2$ .

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Chapter 16.2, Problem 45E

Chapter 16.2, Problem 47E

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Chapter 16 Solutions

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The volume of the region bounded by z = 1 − y 2 and z = y 2 − 1 for 0 ≤ x ≤ 2 .

Solution Summary: The author explains that the volume of the region bounded by z = 1 y 2 and x = 2 is approximately 5.33.

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The volume of the region bounded by $z = 1 - y^{2}$ and $z = y^{2} - 1$ for $0 \leq x \leq 2$ .

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Chapter 16 Solutions

The volume of the region bounded by z = 1 − y 2 and z = y 2 − 1 for 0 ≤ x ≤ 2 .

Solution Summary: The author explains that the volume of the region bounded by z = 1 y 2 and x = 2 is approximately 5.33.

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The volume of the region bounded by z=1−y2 and z=y2−1 for 0≤x≤2.

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Chapter 16 Solutions

The volume of the region bounded by $z = 1 - y^{2}$ and $z = y^{2} - 1$ for $0 \leq x \leq 2$ .