Explanation Given: The provided function is f ( x , y ) = ( 3 x + 2 y ) 2 2 y − x . The region R by parallelogram with vertices ( 2 , 5 ) , ( 4 , 2 ) , ( 0 , 0 ) and ( − 2 , 3 ) . Formula used: Jacobians formula, ∂ ( x , y ) ∂ ( u , v ) = | ∂ x ∂ v ∂ x ∂ u ∂ y ∂ u ∂ y ∂ v | And change of variables for double integrals is given as ∬ R f ( x , y ) d x d y = ∬ S f ( g ( u , v ) , h ( u , v ) ) | ∂ ( x , y ) ∂ ( u , v ) | d u d v The slope-intercept form of equation of line y = m x + c , where ‘ m ’ is the slope of the line and m = y 2 − y 1 x 2 − x 1 . Calculation: The region R by parallelogram with vertices ( 2 , 5 ) , ( 4 , 2 ) , ( 0 , 0 ) and ( − 2 , 3 ) . By using given coordinates, plot the parallelogram. The slope of the straight line between the point ( 0 , 0 ) and ( − 2 , 3 ) is calculated as, m = 3 − 0 − 2 − 0 = − 3 2 The straight-line equation between the point ( 0 , 0 ) and ( − 2 , 3 ) is calculated as, y − 0 = − 3 2 ( x − 0 ) 2 y = − 3 x 3 x + 2 y = 0 The slope of straight line between the point ( 0 , 0 ) and ( 4 , 2 ) is calculated as, m = 2 − 0 4 − 0 = 1 2 The straight-line equation between the point ( 0 , 0 ) and ( 4 , 2 ) is calculated as, y − 0 = 1 2 ( x − 0 ) 2 y = x 2 y − x = 0 The slope of straight line between the point ( 2 , 5 ) and ( 4 , 2 ) is calculated as, m = 2 − 5 4 − 2 = − 3 2 The straight-line equation between the point ( 2 , 5 ) and ( 4 , 2 ) is calculated as, y − 2 = − 3 2 ( x − 4 ) 2 y − 4 = − 3 x + 12 3 x + 2 y = 16 The slope of straight line between the point ( 2 , 5 ) and ( − 2 , 3 ) is calculated as, m = 3 − 5 − 2 − 2 = − 2 − 4 = 1 2 The straight-line equation between the point ( 2 , 5 ) and ( − 2 , 3 ) is calculated as, y − 3 = 1 2 ( x + 2 ) 2 y − 6 = x + 2 2 y − x = 8 Assume that 2 y − x = u and 3 x + 2 y = v . Subtracting these two equations, we get, 3 x + x = v − u x = v − u 4 Put x = v − u 4 in the second equation, 2 y − ( v − u 4 ) = u 2 y = v + 3 u 4 y = v + 3 u 8 The line of the equation 2 y − x = 0 in new coordinate ( u , v ) is given as, 2 y − x = 0 u = 0 The line of the equation 3 x + 2 y = 0 in new coordinate ( u , v ) is given as, 3 x + 2 y = 0 v = 0 The line of the equation x − 2 y = − 8 in new coordinate ( u , v ) is given as, 2 y − x = 8 u = 8 The line of the equation 3 x + 2 y = 0 in new coordinate ( u , v ) is given as, 3 x + 2 y = 16 v = 16 The straight line in a new rectangular co-ordinate is shown as below

10th Edition

ISBN: 9780100453777

Author: Larson

Publisher: YUZU

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Chapter 14.8, Problem 22E

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To calculate: The volume of the provided solid region lying below the surface z=(3x+2y)22y−x and above the plane region R.

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Chapter 14.8, Problem 21E

Chapter 14.8, Problem 23E

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Chapter 14 Solutions

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The volume of the provided solid region lying below the surface z = ( 3 x + 2 y ) 2 2 y − x and above the plane region R .

Solution Summary: The author calculates the volume of the provided solid region lying below the surface and above the plane region R.

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To calculate: The volume of the provided solid region lying below the surface z=(3x+2y)22y−x and above the plane region R.

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Chapter 14 Solutions