   Chapter 15.9, Problem 28E

Chapter
Section
Textbook Problem

Let f be continuous oil [0, 1] and let R be the triangular region with vertices (0, 0), (1, 0), and (0, 1). Show that ∬ R f (x + y) dA = ∫ 0 1 u f ( u )   d u

To determine

To show: The integral Rf(x+y)dA=01uf(u)du.

Explanation

Given:

The function f is continuous on [0,1] and R be the region with vertices (0,0),(1,0) and (0,1).

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)

Proof:

Let u=x+y and v=xy

Find the partial derivative of x  and y with respect to u and v. u=x+y then ux=1 and uy=1 and v=xy then vx=1 and vy=1.

T1=(u,v)(x,y)=|uxuyvxvy| (2)

Substitute the corresponding values in equation (2),

T1=|1111|=1(1)1(1)=11=2

This is in the form of inverse of Jacobian transformation (T1) but obtain Jacobian transformation (T). Therefore the value of Jacobian 2  will be 12

Substitute the given vertices of given region in u=x+y and v=xy to get the image of the region with vertices that is (0,0),(1,0) and (1,1)

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