Explanation Given : f ( x , y ) = x + y , the domain D = { ( x , y ) : x 2 + y 2 ≤ 4 , y ≥ 0 } Calculation : Step 1: The semicircle can be described as a vertically simple region, by the following inequalities: − 2 ≤ x ≤ 2 , 0 ≤ y ≤ 4 − x 2 Step 2: The double integral is evaluated by the following iterated integral: ∫ ∫ D ( x + y ) d A = ∫ − 2 2 ∫ 0 4 − x 2 ( x + y ) d y d x = ∫ − 2 2 x y + 1 2 y 2 | y = 0 4 − x 2 d x = ∫ − 2 2 ( x 4 − x 2 + 1 2 ( 4 − x 2 ) 2 ) d x =

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ISBN: 9781319055844

Author: FRANZOSA

Publisher: VST

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Chapter 16.2, Problem 14E

To determine

The value of the double integral of $f (x, y) = x + y$ over the domain $D = {(x, y) : x^{2} + y^{2} \leq 4, y \geq 0}$ .

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Chapter 16.2, Problem 13E

Chapter 16.2, Problem 15E

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Chapter 16 Solutions

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The value of the double integral of f ( x , y ) = x + y over the domain D = { ( x , y ) : x 2 + y 2 ≤ 4 , y ≥ 0 } .

Solution Summary: The author calculates the value of the double integral of f (x, y) = x+y over the domain D = 4. The semicircle can be described as a vertically simple region.

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The value of the double integral of $f (x, y) = x + y$ over the domain $D = {(x, y) : x^{2} + y^{2} \leq 4, y \geq 0}$ .

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Chapter 16 Solutions

The value of the double integral of f ( x , y ) = x + y over the domain D = { ( x , y ) : x 2 + y 2 ≤ 4 , y ≥ 0 } .

Solution Summary: The author calculates the value of the double integral of f (x, y) = x+y over the domain D = 4. The semicircle can be described as a vertically simple region.

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The value of the double integral of f(x,y)=x+y over the domain D={(x,y):x2+y2≤4,y≥0}.

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Chapter 16 Solutions

The value of the double integral of $f (x, y) = x + y$ over the domain $D = {(x, y) : x^{2} + y^{2} \leq 4, y \geq 0}$ .