Suppose I have a region of integration boded by $x+y \le 8$ and $0 \le y \le x$. I have graphed the bounds, and they form a right triangle with its hypotenuse following the $x$-axis from $[0,8]$ and the right angle at $(4,4)$.
For clarity, the region I am looking at is the intersection of the red and green regions.
I am having trouble figuring out what the bounds should be. I was thinking of splitting up the double integral into two double integrals.
$\int^4_{0} \int^{y=x}_{y=0} f(x) dy dx + \int^8_{4} \int^{y=8-x}_{y=0} f(x) dy dx$
Is this correct?
Best Answer
It is almost correct. For the second integral we still have the condition that $y\leq 8-x$. So the integral is
$$\int^4_{0} \int^{x}_{0} f(x) \, dy \, dx + \int^8_{4} \int^{8-x}_{0} f(x) \, dy \, dx$$
The line $8-x$ is bounded by $4\leq x\leq 8$ ($\color{green}{\textrm{green line}}$). So we get the right half of the triangle area.