Why does coninuity of f important when changing variables from (x,y) -> (u,v))

[zm500]

Homework Statement

Let f be continuous on [0,1] and let R be the triangular region with vertices (0,0), (0,1), and (1,0). Show that
double integrals of f(x+y) w.r.t dA over region R equals single integral of uf(u) w.r.t du from 0 to 1.

Homework Equations

jacobian method

The Attempt at a Solution

i made u = x+y but I can't solve a determinant of just a row.

 

[vela]

Come up with a second variable v.

 

[zm500]

Come up with a second variable v.

Where would I start?

 

[vela]

Try plotting lines of constant u in the xy-plane. A convenient choice for v would be such that lines of constant v are perpendicular to those.

 

[zm500]

Try plotting lines of constant u in the xy-plane. A convenient choice for v would be such that lines of constant v are perpendicular to those.

How can I plot lines of constant u i thought you can only do that uv plane.

 

[vela]

u=constant => x+y=constant

 

[HallsofIvy]

As the others have said, because this is a two dimensional problem, you must have two variables. Take u= x+ y, v= y, for example. That will still give a double integral but now, since f is a function of u only, the first integral, with respect to v, can be done giving "uf(u)du".

 

[zm500]

As the others have said, because this is a two dimensional problem, you must have two variables. Take u= x+ y, v= y, for example. That will still give a double integral but now, since f is a function of u only, the first integral, with respect to v, can be done giving "uf(u)du".

So, the bounds for v are 0\leqv\lequ? I don't understand how they got they got the bounds for u and v. I understand we need an expression for v but I don't know how to come up with that or where to start since u = x+y. Also, are we supposed to come up with an expression visually for v by looking at the region R with vertices (0,0), (1,0), and (0,1).

 

[vela]

That's why I suggested you plot lines of constant u in the xy-plane. It's pretty clear from that what the limits of u are. For instance, the line u=1, which in terms of x and y is x+y=1, runs right along the hypotenuse of the triangular region.

If you choose v so that lines of constant v are perpendicular to lines of constant u (in the xy-plane), you're just rotating the coordinate system, and you'd determine the limits for v the way you usually do.

 

[HallsofIvy]

You are integrating over a triangle, in the xy-plane, with vertices at (0, 0), (1, 0), and (0, 1). The boundaries are the lines y= 0 (joining (0, 0) and (1, 0)), x= 0 (joining (0, 0) and (0, 1)), and x+ y= 1 (joining (1, 0) and (0, 1)). If you let u= x+ y and v= y, then u= x+ v so x= u- v, y= v.

in the uv-plane, the line x= 0 becomes u- v= 0 or v= u. The line y= 0 becomes v= 0 and the line x+y= 1 becomes u= 1. That is a triangle with vertices at (0, 0), (1, 0), and (1, 1).
\int_{x=0}^1\int_{y= 0}^{1- x} f(x+y)dydx
becomes
\int_{u= 0}^1\int_{v= 0}^u f(u)dvdu

By the way- you titled this thread "Why does coninuity of f important when changing variables from (x,y) -> (u,v))". The continuity of f has nothing to do with changing variables- it just guarentees that f is integrable.