on inverse image of a set in R^2

Question

I tried to evaluate a certain double integral by change of variables. When I do that I need to find the preimage of the triangular region bounded by the lines $x=0$, $y=2$ and $x=y$ under the map $x=u(1+v)$, $y=v(1+u)$. To do so I tried to find the pre image of the sides at a time but I am confused in the case of y=2 and x=y. If I can solve the system for u and v interms of x and y,that will be easy but I can't. Please provide your sugestion.

Answer 1

Note that

• if $x = 0$, then $u(1+v) = 0$, so either $u = 0$ or $v = -1$. So the line $x = 0$ becomes the union of the pair of lines $u = 0$ and $v = -1$
• if $y = 2$, then $v = \frac 2{1 + u}$, a hyperbola with vertical asymptote $u = -1$ and horizontal asymptote $v = 0$.
• Since $x - y = u - v$, if $x = y$ then $u = v$, a line.

Now you've got to figure out what region between those curves corresponds to your original triangle.