# Double Integrals - Region delimited by triangle in counterclockwise

A triangle with vertices (0,0), (0,1) and (1,2), counterclockwise, delimits a region. Determine an integral over this region of the following expression: $$\int_{\Omega}(x-y)dx+e^{x+y}dy$$ This kind of exercises fit the line integral, but I don't know how to start solving it, I thought I'd use Green's Theorem. The key is to integrate x first, then y. $$\int_{\Omega}(x-y)dx+e^{x+y}dy=\int_{\Omega}\left[\dfrac{\partial}{\partial x}(e^{x+y}) - \dfrac{\partial}{\partial x}(x-y)\right]\,dx\,dy =\int_{\Omega}\left[e^{x+y} +1\right]\,dx\,dy$$

What would the extremes of integration look like? I don't know how to continue

By green’s theorem, we can convert our line integral (which would require 3 integrals to solve) into a single double integral.

So far, your setup is correct, where we have $$\int_{\Omega}\left[e^{x+y} +1\right]\,dx\,dy$$ over the region(in green)

We can set up our double integral bounds then to be $$\int_0^1 \int_{2x}^{x+1} e^{x+y}+1\,dydx$$

Now all that is left is to evaluate the double integral, which should be pretty easy.

to set up bounds, we'll use the arrow and shadow method. Choose the $$y$$ axis first and draw arrows from $$y=-\infty$$ to $$y=\infty$$, and see what function the arrow passes through and exits from.

It passes through the yellow($$2x$$) first, and then exits purple($$x+1$$). Those are our y bounds. Now shine a light from $$y=\pm\infty$$ and see what shadow the region casts on the $$x$$ axis. This is just the interval $$[0, 1]$$. Those are our $$x$$ bounds.

• Could you explain to me how the extremes were defined? And how do I get to the answer, after performing the integrations "$$e^{3}/6− e/2 +5/6.$$
– Skye
Commented Sep 24, 2022 at 18:26
• @Skye tutorial.math.lamar.edu/classes/calciii/digeneralregion.aspx If you're trying to do green's theorem and line integrals without knowing how to set up double integration bounds... then maybe you should review that first 😐 Commented Sep 24, 2022 at 18:37
• thank you so much!
– Skye
Commented Sep 24, 2022 at 18:38
• @Skye ive added it i guess to the post Commented Sep 24, 2022 at 18:42
• also @Skye if this has answered your question feel free to accept it :) Commented Sep 24, 2022 at 18:45