   Chapter 15.9, Problem 15E

Chapter
Section
Textbook Problem

Use the given transformation to evaluate the integral.15. ∫∫R (x – 3y) dA, where R is the triangular region with vertices (0, 0), (2, 1), and (1,2); x = 2u + v, y = u + 2v

To determine

To evaluate: The integral R(x3y)dA.

Explanation

Given:

The triangular region R with vertices (0,0),(2,1) and (1,2) and x=2u+v, y=u+2v.

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=2u+v then xu=2 and xv=1 and y=u+2v then yu=1 and yv=2.

(x,y)(u,v)=|2112|=2(2)(1)(1)=41=3

From the given integral the function is, x3y and substitute the values of x and y.

x3y=(2u+v)3(u+2v)=2u+v3u6v=u5v

Find the boundary by using the given transformation.

Find the line by using line equation xx1x2x1=yy1y2y1.

A corresponding line for the vertices (0,0) and (2,1) is x=2y.

Substitute x and y in the given line,

x=2y2u+v=2(u+2v)2u+v=2u+4vv=0

A corresponding line for the vertices (2,1) and (1,2) is x+y=3.

Substitute x and y in the given line,

x+y=32u+v+2v+u=33u+3v=3u+v=1

A corresponding line for the vertices (0,0) and (1,2) is y=2x.

Substitute x and y in the given line,

y=2xu+2v=2(2u+v)u+2v=4u+2vu=0

In uv plane, the region is defines as follows,

R={(u,v)|0v1u,0u1}

By using equation (1) the integral will be,

R(x3y)dA=0101u(u5v)|3|dvdu=30101u(u+5v)dvdu

Integrate the above integral with respect to v and apply the limit of it

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