Integrate over a set $B = \left\{ (x,y) \in \Bbb R^2: 2\leq x \leq y \leq 6 \right\}$.

Question

How do I integrate over the following set?

$$B = \left\{ (x,y) \in \Bbb R^2: 2\leq x \leq y \leq 6 \right\}$$

This may seem trivial but I really am not sure how to find the bounds. I thought since $x \leq y$ it would indicate an area under a function $y = x$ where $x \in [2,6]$ and so from the graph I could determine $y \in [0, x]$ but I don't think this is correct since $2 \leq y$ so the bounds for $y$ should be $y \in [2,x]$ but I still am not sure whether this is correct. And also isn't it possible to rewrite the set so that I was able to see the bounds straight away?

Answer 1

The condition $B = \left\{ (x,y) \in \Bbb R^2: 2\leq x \leq y \leq 6 \right\}$ ie equivalent to

• $2 \leq y \leq 6$
• $2\leq x \leq y $

then make a sketch of the lines

• $y=2$, $y=6$
• $x=2$, $x=y$

and the region $B$ is given by the the triangle between the lines $x=2$, $y=6$ and $x=y$.

Answer 2

From the bounds, you should note that ${x}$ runs from ${2}$ to ${6}$. From the bounds also you can see that ${y}$ runs from whatever the value of ${x}$ is to the value of ${6}$. In double integral notation, that is

$${\int_{2}^{6}\int_{x}^{6}f(x,y)dydx}$$

The ${dy}$ being first is very important! The constraints between ${x}$ and ${y}$ are coupled, so you cannot just change the order of integration. If you did, however, want to change the order of integration - you can make a sketch of the area in ${\mathbb{R}^2}$ and you should be able to find the new constraints. Or equivalently, from the bounds - ${y}$ runs from ${2}$ to ${6}$, and this would force ${x}$ to run from ${2}$ to whatever the value of $y$ is. That is,

$${\int_{2}^{6}\int_{2}^{y}f(x,y)dxdy}$$

Upon request, I can go through the graphical approach of this problem