Change of Variables of f(x+y) (In Multiple Integrals)

[Jadehaan]

Homework Statement

The problem is as follows: Let T be the triangle with vertices (0,1), (1,0), (0,0). Compute the integral \int\int\frac{sin^{2}(x+y)}{(x+y)} dxdy by making an appropriate change of variables. (Hint: check #24 Section 15.9)

Homework Equations

Problem 24 in 15.9 of Stewart Calculus Early Transcendentals: Let f be continuous on [0,1] and let R be the triangular region with vertices (0,1), (1,0) and (0,0). Show that
\int\intf(x+y)dA =\int_{0}^{1}uf(u)du

The Attempt at a Solution

I am confused at what values I should assign u and v in order to change the variables appropriately. Assuming the answer to #24, I obtained the solution to be (1/2)-(1/4)sin(2)
Thanks for any help or tips that point me in the right direction,
Jim

 

[LCKurtz]

The key to #24 is the fact that the side of the triangle not on one of the axes has the equation x + y = 1. That suggests letting one of the variables by x + y. And there is no reason to change the other variable, so try:

u = x + y
v = x

The corners of the xy triangle go to (u,v) like this:

(0,0 ) --> (0,0)
(1,0) --> (1,1)
(0,1) --> (1,0)

Your exercise is to show that

\int\int_A f(x+y)\,dA = \int_0^1uf(u)\,du

Notice on the right side, what would normally be a double integral has been integrated in the v variable (the one that isn't in the f()) already, so set up the change of variables that way. The way we set it up, the non-x+y variable is v also. Since |J| = 1 I will leave the Jacobian out. Our vu limits become:

\int_0^1\int_0^u f(u)\,dvdu

Now you can do the inner integral without knowing f.

 

[Jadehaan]

Got it, thanks!