Let $0<a<b$ and consider the triangular region bounded by the three points $(a,a)$, $(b,a)$ and $(a,b)$.
If we would integrate some function $F(x,y)$ over this region, how does one write down the bounds of integration here?
$$
\int_{\mathrm{Triangle}} F(x,y) \ dx dy = ?
$$
The triangle is bounded by the line $y(x) = – x + (a+b)$ in the $xy$-plane, and so this makes me think that it should be
$$
\int_{\mathrm{Triangle}} F(x,y) \ dx dy = \int_{a}^b \int_{a}^{-x+(a+b)} F(x,y) \ dy dx
$$
But I don't think this is correct because picking $F(x,y) = 1$ does not give me the area $ab/2$ which one would expect.
What am I doing wrong?
Best Answer
You are doing nothing wrong except assuming that the area of the triangle is $\cfrac{ab}{2}$.
Vertices are $A (a, a), B (b, a), C (a, b)$
$AB = b - a, AC = b - a$ and $\angle ABC = 90^0$ so the area of the triangle is $\cfrac{1}{2} (b-a)^2$.
Also, $ \displaystyle \int_{a}^b \int_{a}^{-x+(a+b)} 1 \ dy \ dx = \cfrac{1}{2} (b-a)^2$