How to determine the integration boundaries of the following double integral? Calculate the following double integral:
$\int\limits$$\int\limits_T$  $[xsen(x) + ysen(x+y)]$ $dxdy$
Where the region $T$ is the triangle of vertices $(1,0)$, $(0,1)$ y $(3,3)$.
To could determine the boundaries of the integral  I did the following:


*

*I made a graph of the triangle

*I found the equations that describe the three lines of the triangle:


*

*The one which goes from $(0,1)$ to  $(3,3)$ is $y=2/3x+1$

*The one which goes from $(0,1)$ to  $(1,0)$ is $y=-x+1$

*The one which goes from $(1,0)$ to  $(3,3)$ is $y=3/2x-3/2$



Now, I'm stuck from here. I don't know how to establish the boundaries for $x$ and $y$ given the restriction named $T$. Any hint?
 A: You can split your integral as the sum of the red and blue surface:
\begin{align}
\text{Red}&=\int_0^1\int_{-x+1}^{2/3x+1}(x\sin(x)+y\sin(x+y))\,\mathrm dy\mathrm dx\\
\text{Blue}&=\int_1^3\int_{3/2x-3/2}^{2/3x+1}(x\sin(x)+y\sin(x+y))\,\mathrm dy\mathrm dx
\end{align}

A: We have the picture

Equation of line $\overset{\leftrightarrow}{AC}$: 
$$
\frac{3-1}{3-0}=\frac{y-1}{x-0}\implies y=x+\frac{3}{2}
$$
Equation of line $\overset{\leftrightarrow}{RC}$: 
$$
\frac{3-0}{3-1}=\frac{y-0}{x-1}\implies y=\frac{3}{2}x-\frac{3}{2}
$$
Then 
\begin{align}
\displaystyle\iint_{T}[xsen(x) + ysen(x+y)] \mathrm{d} A
=&
\displaystyle\iint_{\substack{0\leq x \leq 1 \\ 0\leq y\leq x+\frac{3}{2}}}[xsen(x) + ysen(x+y)] \mathrm{d} A
\\
&+
\displaystyle\iint_{\substack{1\leq x \leq 3 \\ \frac{3}{2}x-\frac{3}{2}\leq y\leq x+\frac{3}{2}}}[xsen(x) + ysen(x+y)] \mathrm{d} A
\\
=&
\displaystyle \int_{0}^{1}\int_{0}^{x+\frac{3}{2}}[xsen(x) + ysen(x+y)] \mathrm{d} y\mathrm{d}x
\\
&+
\displaystyle\int_{1}^{3}\int_{\frac{3}{2}x-\frac{3}{2}}^{x+\frac{3}{2}}[xsen(x) + ysen(x+y)] \mathrm{d} y\mathrm{d}x
\end{align}
