Explanation Given: The function f ( x , y ) = ( x − y ) ( x + 4 y ) R: Region bounded by the parallelogram with coordinates, ( x , y ) = ( 0 , 0 ) ( x , y ) = ( 1 , 1 ) ( x , y ) = ( 5 , 0 ) ( x , y ) = ( 4 , − 1 ) Formula used: Using Jacobians formula δ ( x , y ) δ ( u , v ) = | δ x δ u δ x δ v δ y δ u δ y δ v | And, change of variables for double integrals ∬ R f ( x , y ) d x d y = ∬ S f ( g ( u , v ) , h ( u , v ) ) | δ ( x , y ) δ ( u , v ) | d u d v Using 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: Let us assume x = u + 4 v 5 and y = u − v 5 Which gives us the value of u and v as, u = x + 4 y ⋯ ( Α ) v = x − y ⋯ ( B ) Calculate the Jacobian as, δ ( x , y ) δ ( u , v ) = | δ x δ u δ x δ v δ y δ u δ y δ v | δ ( x , y ) δ ( u , v ) = | 1 5 4 5 1 5 − 1 5 | δ ( x , y ) δ ( u , v ) = − 1 5 Construct a graph with the help of the given conditions using the slope-intercept form of equation of line y = m x + c

Calculus: Early Transcendental Functions (MindTap Course List)

6th Edition

ISBN: 9781285774770

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 14.8, Problem 25E

To determine

To calculate: The volume lying below the surface f(x,y)=(x−y)(x+4y) and above the plane region R using change of variables where region R is bounded by the parallelogram with the following coordinates

(x,y)=(0,0)(x,y)=(1,1)(x,y)=(5,0)(x,y)=(4,−1)

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Chapter 14.8, Problem 24E

Chapter 14.8, Problem 26E

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Chapter 14 Solutions

Calculus: Early Transcendental Functions (MindTap Course List)

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - Prob. 1E Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy

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The volume lying below the surface f ( x , y ) = ( x − y ) ( x + 4 y ) and above the plane region R using change of variables where region R is bounded by the parallelogram with the following coordinates ( x , y ) = ( 0 , 0 ) ( x , y ) = ( 1 , 1 ) ( x , y ) = ( 5 , 0 ) ( x , y ) = ( 4 , − 1 )

Solution Summary: The author explains how to calculate the volume below the surface and above the plane region R using change of variables.

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To calculate: The volume lying below the surface f(x,y)=(x−y)(x+4y) and above the plane region R using change of variables where region R is bounded by the parallelogram with the following coordinates

(x,y)=(0,0)(x,y)=(1,1)(x,y)=(5,0)(x,y)=(4,−1)

Want to see the full answer?

Chapter 14 Solutions

The volume lying below the surface f ( x , y ) = ( x − y ) ( x + 4 y ) and above the plane region R using change of variables where region R is bounded by the parallelogram with the following coordinates ( x , y ) = ( 0 , 0 ) ( x , y ) = ( 1 , 1 ) ( x , y ) = ( 5 , 0 ) ( x , y ) = ( 4 , − 1 )

Solution Summary: The author explains how to calculate the volume below the surface and above the plane region R using change of variables.

Concept explainers

Videos

To calculate: The volume lying below the surface f(x,y)=(x−y)(x+4y) and above the plane region R using change of variables where region R is bounded by the parallelogram with the following coordinates (x,y)=(0,0)(x,y)=(1,1)(x,y)=(5,0)(x,y)=(4,−1)

Want to see the full answer?

Chapter 14 Solutions

To calculate: The volume lying below the surface f(x,y)=(x−y)(x+4y) and above the plane region R using change of variables where region R is bounded by the parallelogram with the following coordinates

(x,y)=(0,0)(x,y)=(1,1)(x,y)=(5,0)(x,y)=(4,−1)