Limits of $u$ and $v$ I am trying to evaluate a double integral $$I=\int_{0}^2\int_{0}^{2-x}(x+y)^2e^{\frac{2y}{x+y}}dydx$$
I used the transformation $$x+y=v, y=uv$$
That is $$x=v(1-u), y=uv$$
We get the Jacobian as:
$$J=\left|\begin{array}{ll}
\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\
\frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}
\end{array}\right|=\left|\begin{array}{cc}
-v & 1-u \\
v & u
\end{array}\right|=-v$$
So we get:
$$I=\iint_{D(u,v)}v^3e^{2u}dudv$$
I am unable to figure out the limits of $u$ and $v$
 A: As per your transformation, we have $u = \frac{y}{x+y}, v = x + y$
Note that $u$ is not defined at point $(x = 0, y = 0)$, but that is just a point on the boundary of the original region and we can still transform the region and evaluate the integral. Keeping $(0, 0)$ aside, the original region can be expressed as $0 \lt x + y \lt 2, x, y \gt 0$ and that translates to $~0 \lt \frac{y}{x+y} \lt 1,$ i.e. $~0 \lt u \lt 1$.
So $0 \lt u \lt 1, 0 \lt v \lt 2$ are the bounds of the new integral.
Alternatively, use the transformation
$u = y, v = x + y$. As $x, y \gt 0$, we must have $u \lt v$ so bounds are simply $0 \lt u \lt v, 0 \lt v \lt 2$. $|J| = 1$.
The integral becomes,
$ \displaystyle \int_0^2 \left[\int_0^v (v^2  e^{2u / v}) ~du \right]dv = \frac{e^2 - 1}{2} \int_0^2 v^3 ~dv$
A: One has
\begin{align*}
I&=\int_{0}^2\int_{0}^{2-x}(x+y)^2e^{\frac{2y}{x+y}}dy\ dx \\
&= \int_0^2 \int_{1}^{x/2} -\left( \frac{x}{u}\right)^2 e^{2-2u} \frac{x}{u^2} du \ dx \quad \quad (\text{substituting } y \text{ by }u=\frac{x}{x+y})\\
&= \int_0^2 \int_{x/2}^{1}  \frac{x^3}{u^4} e^{2-2u}  du \ dx \\
&=\int_0^1 \int_{0}^{2u}  \frac{x^3}{u^4} e^{2-2u}  dx \ du \quad \quad (\text{Fubini})\\
&=\int_0^1 \frac{e^{2-2u}}{u^4} \times \frac{(2u)^4}{4} du\\
&=\int_0^1 4e^{2-2u}  du\\
&\boxed{= 2(e^2-1)}
\end{align*}
A: Map the region defined by $D(x,y) \to D'(u,v)$. One should look at the boundary and the interior region.
For example - the boundary defined by $x + y = 2$ implies that $v = 2$ and $u = y/2$ and since $y \in [0,2]$ this is a straight line segment $v = 2$ that runs from $u \in [0,1]$.
Check the other boundaries and you should obtain the box in $u-v$ given by $[0,1] \times [0,2]$.
