How to determine the integration boundaries of the following double integral?

Question

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:

1. I made a graph of the triangle
2. 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?

Answer 1

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}

Answer 2

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}