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Suppose I have to integrate $$\int_T f(x, y) \text{d}x\text{d}y$$

where $T$ is a triangle, and the vertexes are given. How do you construct the bounds for the integral? I find myself stuck every time due to some lack of clarity.

For example, say $T$ is the triangle with vertices $(0, 0)$, $(0, 1)$ and $(1, 1)$.

For sure, I would draw a picture on the cartesian plane, otherwise I would find it like impossible to understand the bonds mentally. Yet after having drawn them, I would still have doubts.

I know, in the case above, that $x$ runs from $0$ to $1$. In the same way, I understand that $y$ follows the $y = x$ paths, but here to say what are the correct bonds for $y$?

I find two possibilities: $y \in (1, x)$ or $y\in (x, 1)$, because when $x = 0$ then $y = 1$ (it's a vertex) but also when $x = 1$ we have $y = 1$.

I'm so confused!

Is there a general way to reason about those problems in order to understand well the path and the order of the bounds?

Thank you!

Kenny Wong
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Heidegger
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1 Answers1

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I think it will help to first write your triangle $T$ in set notation as follows: $$ T=\{ (x,y)\in\mathbb R^2\colon 0\leq x\leq y\leq 1\}. $$ Try to make a drawing and see, that this is correct. So $(x,y)\in T$ if and only if $0\leq x\leq 1$ and $x\leq y\leq 1$ or equivalently $0\leq y\leq 1$ and $0\leq x\leq y$. Thus, $$ \int_T f(x,y)\,dx dy=\int_0^1 \int_x^1 f(x,y)\, dy dx=\int_0^1 \int_0^y f(x,y)\, dx dy. $$

stochastic-conch
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  • This is for the particular case ; for the general case, you should push your explanation into a more operational answer : what do we do practically when we are given the$(x,y)$ coordinates of the 3 vertices ? – Jean Marie May 10 '23 at 10:36