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 = ∫

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Question

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

Chapter 16 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 16.1 - Prob. 1PQ Ch. 16.1 - Prob. 2PQ Ch. 16.1 - Prob. 3PQ Ch. 16.1 - Prob. 4PQ Ch. 16.1 - Prob. 5PQ Ch. 16.1 - Prob. 6PQ Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E

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Prob. 2E Ch. 16.5 - Prob. 3E Ch. 16.5 - Prob. 4E Ch. 16.5 - Prob. 5E Ch. 16.5 - Prob. 6E Ch. 16.5 - Prob. 7E Ch. 16.5 - Prob. 8E Ch. 16.5 - Prob. 9E Ch. 16.5 - Prob. 10E Ch. 16.5 - Prob. 11E Ch. 16.5 - Prob. 12E Ch. 16.5 - Prob. 13E Ch. 16.5 - Prob. 14E Ch. 16.5 - Prob. 15E Ch. 16.5 - Prob. 16E Ch. 16.5 - Prob. 17E Ch. 16.5 - Prob. 18E Ch. 16.5 - Prob. 19E Ch. 16.5 - Prob. 20E Ch. 16.5 - Prob. 21E Ch. 16.5 - Prob. 22E Ch. 16.5 - Prob. 23E Ch. 16.5 - Prob. 24E Ch. 16.5 - Prob. 25E Ch. 16.5 - Prob. 26E Ch. 16.5 - Prob. 27E Ch. 16.5 - Prob. 28E Ch. 16.5 - Prob. 29E Ch. 16.5 - Prob. 30E Ch. 16.5 - Prob. 31E Ch. 16.5 - Prob. 32E Ch. 16.5 - Prob. 33E Ch. 16.5 - Prob. 34E Ch. 16.5 - Prob. 35E Ch. 16.5 - Prob. 36E Ch. 16.5 - Prob. 37E Ch. 16.5 - Prob. 38E Ch. 16.5 - Prob. 39E Ch. 16.5 - Prob. 40E Ch. 16.5 - Prob. 41E Ch. 16.5 - Prob. 42E Ch. 16.5 - Prob. 43E Ch. 16.5 - Prob. 44E Ch. 16.5 - Prob. 45E Ch. 16.5 - Prob. 46E Ch. 16.5 - Prob. 47E Ch. 16.5 - Prob. 48E Ch. 16.5 - Prob. 49E Ch. 16.5 - 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Prob. 26CRE Ch. 16 - Prob. 27CRE Ch. 16 - Prob. 28CRE Ch. 16 - Prob. 29CRE Ch. 16 - Prob. 30CRE Ch. 16 - Prob. 31CRE Ch. 16 - Prob. 32CRE Ch. 16 - Prob. 33CRE Ch. 16 - Prob. 34CRE Ch. 16 - Prob. 35CRE Ch. 16 - Prob. 36CRE Ch. 16 - Prob. 37CRE Ch. 16 - Prob. 38CRE Ch. 16 - Prob. 39CRE Ch. 16 - Prob. 40CRE Ch. 16 - Prob. 41CRE Ch. 16 - Prob. 42CRE Ch. 16 - Prob. 43CRE Ch. 16 - Prob. 44CRE Ch. 16 - Prob. 45CRE Ch. 16 - Prob. 46CRE Ch. 16 - Prob. 47CRE Ch. 16 - Prob. 48CRE Ch. 16 - Prob. 49CRE Ch. 16 - Prob. 50CRE Ch. 16 - Prob. 51CRE Ch. 16 - Prob. 52CRE Ch. 16 - Prob. 53CRE Ch. 16 - Prob. 54CRE Ch. 16 - Prob. 55CRE Ch. 16 - Prob. 56CRE Ch. 16 - Prob. 57CRE Ch. 16 - Prob. 58CRE Ch. 16 - Prob. 59CRE Ch. 16 - Prob. 60CRE Ch. 16 - Prob. 61CRE

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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 $0 \leq x \leq 2$ .