   Chapter 15.9, Problem 16E

Chapter
Section
Textbook Problem

Use the given transformation to evaluate the integral.16. ∫∫R (4x – 8y) dA, where R is the parallelogram with vertices (–1, 3), (1, –3), (3, -1), and (1, 5); x = 1 4 (u + v), y = 1 4 (v – 3u)

To determine

To evaluate: The integral R(4x+8y)dA.

Explanation

Given:

The parallelogram region R with vertices (1,3),(1,3),(3,1) and (1,5) and x=14(u+v), y=14(v3u).

Property used: Change of Variable

Change of Variable in double integral is given by,

Rf(x,y)dA=Sf(x(u,v),y(u,v))|(x,y)(u,v)|dudv (1)

Calculation:

Obtain the Jacobian, (x,y)(u,v)=|xuxvyuyv|

Find the partial derivative of x and y with respect to u and v. x=14(u+v) then xu=14 and xv=14 and y=14(v3u) then yu=34 and yv=14.

(x,y)(u,v)=|14143414|=14(14)(14)(34)=116+316=416

On further simplification find the value of Jacobian.

(x,y)(u,v)=14

From the given integral the function is, 4x+8y and substitute the values of x and y.

4x+8y=414(u+v)+814(v3u)=u+v+2v6u=3v5u

Find the boundary by using the given transformation.

Find the line by using line equation xx1x2x1=yy1y2y1.

A corresponding line for the vertices (1,3) and (1,3) is 3x+y=0.

Substitute x and y in the given line,

3x+y=0314(u+v)+14(v3u)=034u+34v+14v34u=044v=0v=0

The corresponding line for the vertices (1,3) and (3,1) is xy=4.

Substitute x and y in the given line,

xy=414(u+v)14(v3u)=414u+14v14v+34u=444u=4u=4

A corresponding line for the vertices (3,1) and (1,5) is 3x+y=8

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