Given D={(x,y)|x⩾0,y⩾0,x+y⩽1}, find the value of $\iint _{D} e^{x+y} dxdy$
Method 1: \begin{array}{l} \iint _{D} e^{x+y} dxdy=\int _{0}^{1}\int _{0}^{1-y} e^{x+y} dxdy\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _{0}^{1} e^{y}\left( e^{1-y} -1\right) dy\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1 \end{array}
Then I tried to solve this question by Jacobian
\begin{array}{l} Let\ u=x+y,\ v=x,\ 0\leqslant v\leqslant u\leqslant 1\\ \begin{cases} x=v\\ y=u-v \end{cases}\\ J=\begin{vmatrix} x_{u} & y_{u}\\ x_{v} & y_{v} \end{vmatrix} =\begin{vmatrix} 0 & 1\\ 1 & -1 \end{vmatrix} =-1 \end{array}
The different results that came out from the different integral bounds confused me. \begin{cases} \int _{0}^{1}\int _{0}^{u} e^{u} |J|dvdu=1\\ \int _{0}^{1}\int _{0}^{1} e^{u} |J|dvdu=e-1 \end{cases}
The integral bounds of Jacobian should be constants, but the result isn't the same as the one in method 1.
And I tried to change the integral bounds from $\int _{0}^{1}\int _{0}^{1} dvdu$ to $\int _{0}^{1}\int _{0}^{u} dvdu$. The results are the same, but the integral bounds of Jacobian aren't like that.
Please help. Thanks in advance.
The bounds are not required to be constants. They are just preferred to be nice to work with.
$$\begin{align}\iint_D \mathrm e^{x+y}\,\mathrm d x\,\mathrm d y&=\iint_{0\leqslant x\leqslant x+y\leqslant 1}\mathrm e^{x+y}\,\mathrm d x\,\mathrm d y\\&= \iint_{0\leq v\leqslant u\leqslant 1} \mathrm e^u\,\lVert\mathbf J_{v,u}^{v,u-v}\rVert\,\mathrm d v\,\mathrm d u\\&=\int_0^1\mathrm e^u\int_0^\color{crimson}u 1\,\mathrm d v\,\mathrm d u\\&=1\end{align}$$