Explanation Given: Let G ( u , v ) = ( u − 2 v , v ) . Formula Used: Jac ( G ) = | ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v | ... ( 1 ) If G : D 0 → D and if f is continuous then, ∬ D f ( x , y ) d x d y = ∬ D 0 f ( x ( u , v ) , y ( u , v ) ) | ∂ ( x , y ) ∂ ( u , v ) | d u d v .... ( 2 ) Calculation: Given mapping is: G ( u , v ) = ( u − 2 v , v ) The given region in xy-plane is bounded by: x + 2 y = 6 , x + 2 y = 10 , y = 1 and y = 3 We have to find ∬ D ( x + 3 y ) d x d y . So, the transformations will be: x + 2 y = u , y = v → x = u − 2 v Therefore, G ( u , v ) = ( u − 2 v , v ) , 6 ≤ u ≤ 10 , 1 ≤ v ≤ 3 To calculate the area of G ( R ) , first we need to find the Jacobian by using the formula(1): x = u − 2 v ∂ x ∂ u = 1 , ∂ x ∂ v = − 2 And y = v ∂ y ∂ u = 0 , ∂ y ∂ v ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.6, Problem 31E

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To calculate:

$\iint_{D} (x + 3 y) d x d y$ ,where D is the shaded region in the givenfigure.

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Chapter 16.6, Problem 30E

Chapter 16.6, Problem 32E

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∬ D ( x + 3 y ) d x d y ,where D is the shaded region in the givenfigure.

Solution Summary: The author explains how to calculate the area of G ( R ) by finding the Jacobian.

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$\iint_{D} (x + 3 y) d x d y$ ,where D is the shaded region in the givenfigure.

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Chapter 16 Solutions

∬ D ( x + 3 y ) d x d y ,where D is the shaded region in the givenfigure.

Solution Summary: The author explains how to calculate the area of G ( R ) by finding the Jacobian.

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To calculate: ∬D(x+3y)dxdy,where D is the shaded region in the givenfigure.

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Chapter 16 Solutions

To calculate:

$\iint_{D} (x + 3 y) d x d y$ ,where D is the shaded region in the givenfigure.