I know there have been other questions very similar to this, but I can't seem to understand how to determine the limits of integration when changing variables. Take for example
$$\int_0^{\frac{1}{2}} \int_0^{1-2y} e^{x/(x+2y)}dxdy$$
with the change of variables $u = x$ and $v = x+2y$. Getting the Jacobian is simple $(\frac{1}{2})$.
Other answers recommend looking in the $x$–$y$ plane and see how the bounding curves transform in the $u$–$v$ plane. So the limit $1-2y$ corresponds to a line from $(0,\frac{1}{2})$ to $(1,0)$. That is the line
$$x = 1 – 2y \implies x + 2y = v = 1$$ So the bounding region should be the line $v=1$ and we should get a square with vertices at $(0,0)$ and $(1,1)$ and the integral should now be
$$\frac{1}{2}\int_0^1 \int_0^1 e^{u/v} dudv$$ However, according to wolframalpha, this integral does not converge, bu the integral
$$\frac{1}{2}\int_0^1 \int_0^v e^{u/v}dudv $$ does converge and to the correct value.
Why are the correct limits of integration for $u$ from $0$ to $v$ and not $0$ to $1$?
Best Answer
OK, you know from Wolfram Alpha the region of integration is the triangle with vertices $(u,v) = \{(0,0),(0,1),(1,1)\}$.
Let me convince you why a point in its complement in the square $[0,1] \times [0,1]$ (your claimed region of integration) is not part of the region of integration.
Suppose $v < u$. Then $x + 2y < x$. So $2y < 0$ and so $y < 0$. That's not possible with our starting region of integration!