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Find (2x + 3y)dA where R is the parallelogram with vertices (0,0), (-1,-4), (-3,-2), R and (-4,-6). Use the transformation x = - u - 3v, y = = 4u 2v .

Find (2x + 3y)dA where R is the parallelogram with vertices (0,0), (-1,-4), (-3,-2), R and (-4,-6). Use the transformation x = - u - 3v, y = = 4u 2v .

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question

5.7.6

**Problem Statement:** Find the double integral \(\iint_R (2x + 3y) \, dA\) where \(R\) is the parallelogram with vertices \((0,0)\), \((-1,-4)\), \((-3,-2)\), and \((-4,-6)\). Use the transformation \(x = -u - 3v\), \(y = -4u - 2v\). **Explanation:** The task is to calculate the integral of the function \(2x + 3y\) over the region \(R\), which is defined as a parallelogram in the \(xy\)-plane. The region's vertices are given as \((0,0)\), \((-1,-4)\), \((-3,-2)\), and \((-4,-6)\). To solve this problem, we employ a change of variables defined by the transformation: - \(x = -u - 3v\) - \(y = -4u - 2v\) This transformation allows us to potentially simplify the computation of the integral by mapping the parallelogram \(R\) into a new region in the \(uv\)-plane. The task would typically involve finding the Jacobian of the transformation to correctly handle the area differential when changing from \(dA\) to \(dudv\).
Transcribed Image Text:**Problem Statement:** Find the double integral \(\iint_R (2x + 3y) \, dA\) where \(R\) is the parallelogram with vertices \((0,0)\), \((-1,-4)\), \((-3,-2)\), and \((-4,-6)\). Use the transformation \(x = -u - 3v\), \(y = -4u - 2v\). **Explanation:** The task is to calculate the integral of the function \(2x + 3y\) over the region \(R\), which is defined as a parallelogram in the \(xy\)-plane. The region's vertices are given as \((0,0)\), \((-1,-4)\), \((-3,-2)\), and \((-4,-6)\). To solve this problem, we employ a change of variables defined by the transformation: - \(x = -u - 3v\) - \(y = -4u - 2v\) This transformation allows us to potentially simplify the computation of the integral by mapping the parallelogram \(R\) into a new region in the \(uv\)-plane. The task would typically involve finding the Jacobian of the transformation to correctly handle the area differential when changing from \(dA\) to \(dudv\).
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