Finding $\iint_D e^{x+y} \,\mathrm{d}x\,\mathrm{d}y$ without substitution

Question

Is it possible to set this integral up without using substitution?

$$\iint_D e^{x+y} \,\mathrm{d}x\,\mathrm{d}y\,,$$ where

$$D = \left\{-1\le x+y \le 1, -1 \le -x + y \le 1\right\}$$

The answer is: $e-\frac{1}{e}$

Answer 1

Sure, and indeed it is quite easy. First you can divide $D$ in $D_1=\left\{0\leq x\leq 1, x-1 \le y \le -x+1\right\}$ and $D_2=\left\{-1\leq x\leq 0, -x-1 \le y \le x+1\right\}$, then

$$ \iint_{D_1} e^{x+y} dxdy=\int_0^1 \left(\int_{x-1}^{-x+1} e^{x+y} dy\right)dx= \int_0^1[e^{x+y}]_{x-1}^{-x+1} dx=\int_0^1 (e-e^{2x-1}) dx $$

now yo can complete the whole exercise by yourself.

Answer 2

Yes you are integrating over a square bound by $4$ given line segments. Your integral should be

$\displaystyle \int_{-1}^{0}\int_{-1-y}^{1+y} e^{x+y} \, dx \, dy + \displaystyle \int_{0}^{1}\int_{y-1}^{1-y} e^{x+y} \, dx \, dy$

You do not need substitution.