Explanation Given : f ( x , y ) = ( x + y + 1 ) = 2 , the triangle with vertices (0,0),(4,0) and (0,8). Calculation : Step 1: The region D can be described as a vertically simple region, but first we need to find the equation of the line, joining the points (4,0) and (0,8) : y − 8 = 8 − 0 0 − 4 ( x − 0 ) ⇒ y − 8 = − 2 x ⇒ y = 8 − 2 x Step 2: The following inequalities for D are obtained as 0 ≤ x ≤ 4 , 0 ≤ y ≤ 8 − 2 x Step 3: The double integral f ( x , y ) = ( x + y + 1 ) = 2 over the triangle D is evaluated by the following iterated integral: ∫ ∫ D f ( x , y ) d A = ∫ 0 4 ∫ 0 8 − 2 x ( x + y + 1 ) − 2 d y d x = ∫ 0 4 − ( x + y + 1 ) − 1 | y = 0 8 − 2 x d x = ∫ 0 4 ( − ( x + 8 − 2 x + 1 ) − 1 + ( x + 0 + 1 ) − 1 ) d x = ∫ 0 4 ( 1 x + 1 − 1 9 − x ) d x = ln ( x + 1 ) + ln ( 9 − x ) | 0 4 = ln 5 + l n 5 − ( l n 1 + l n 9 ) = 2 l n 5 − ln 9 = ln 25 9 ≈ 1...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 16.2, Problem 13E

To determine

The value of the double integral of $f (x, y) = {(x + y + 1)}^{= 2}$ over the triangle with vertices (0,0) ,(4,0) and (0,8).

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Chapter 16.2, Problem 12E

Chapter 16.2, Problem 14E

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Chapter 16 Solutions

CALCULUS (CLOTH)

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. 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 value of the double integral of f ( x , y ) = ( x + y + 1 ) = 2 over the triangle with vertices (0,0) ,(4,0) and (0,8).

Solution Summary: The author calculates the value of the double integral of f (x, y) = 2 over the triangle with vertices (0,0),(4,0) and (0,8) to be approximately 1.02.

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The value of the double integral of $f (x, y) = {(x + y + 1)}^{= 2}$ over the triangle with vertices (0,0) ,(4,0) and (0,8).

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Chapter 16 Solutions

The value of the double integral of f ( x , y ) = ( x + y + 1 ) = 2 over the triangle with vertices (0,0) ,(4,0) and (0,8).

Solution Summary: The author calculates the value of the double integral of f (x, y) = 2 over the triangle with vertices (0,0),(4,0) and (0,8) to be approximately 1.02.

Videos

The value of the double integral of f(x,y)=(x+y+1)=2 over the triangle with vertices (0,0) ,(4,0) and (0,8).

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Chapter 16 Solutions

The value of the double integral of $f (x, y) = {(x + y + 1)}^{= 2}$ over the triangle with vertices (0,0) ,(4,0) and (0,8).