Explanation Given: The plane z = x + 2 y + 4 and the region D in x y -plane bounded by y = 2 x , y = 3 − x and y = 0 Concept Used: Volume under z = f ( x , y ) is defined as V = ∬ D f ( x , y ) d A A double integral ∬ R f ( x , y ) d A over the range R : a ≤ x ≤ b , g 1 ( x ) ≤ y ≤ g 2 ( x ) is defined as ∬ R f ( x , y ) d A = ∫ a b ∫ g 1 ( x ) g 2 ( x ) f ( x , y ) d y d x And A double integral ∬ R f ( x , y ) d A over the range R : m 1 ( x ) ≤ x ≤ m 2 ( x ) , c ≤ y ≤ d is defined as ∬ R f ( x , y ) d A = ∫ c d ∫ m 1 ( x ) m 2 ( x ) f ( x , y ) d x d y Calculation: For the givenplane z = x + 2 y + 4 and the region D in x y -plane bounded by y = 2 x , y = 3 − x and y = 0 The volume is given by: V = ∬ D ( x + 2 y + 4 ) d A Sketching the region of integration

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 12.2, Problem 52PS

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To find:The volume under the plane $z=x+2y+4$ and above the region $D$ bounded by $y=2x$ , $y=3−x$ and $y=0$

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Chapter 12.2, Problem 51PS

Chapter 12.2, Problem 53PS

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The volume under the plane z = x + 2 y + 4 and above the region D bounded by y = 2 x , y = 3 − x and y = 0

Solution Summary: The author explains that the volume under the plane z=x+2y+4 and above the region D is defined as V=displaystyle undersetDiint

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To find:The volume under the plane $z=x+2y+4$ and above the region $D$ bounded by $y=2x$ , $y=3−x$ and $y=0$

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