The triangle in the $(x,y)$ plane bounded by the lines $y=x, ~y=-x$ and $y=2$

Question

I want to find $\iint_Uf \,d{x}\,d{y} $ where $U$ is the triangle in the $(x,y)$ plane bounded by the lines $y=x, ~y=-x$ and $y=2$ and $f=x+y-2xy$.

My question is how come $x \not \in [-2,2]$? Rather, correct answer is attained by $x \in [0,2]$.

In other words, why is it $\displaystyle \int_0^{2}\int_{-x}^{x}(x+y-2xy)\,dy\,dx $ and not $\displaystyle \int_{-2}^{2}\int_{-x}^{x}(x+y-2xy)\,dy\,dx $

I've graphed it, and the triangle has vertices $[0,0], ~ [-2,2], ~ [2,2]$ but I still can't see it.

Answer 1

The limits of the region are: $$ 0<y<2 \qquad \mbox{and}\qquad -y<x-y $$

so the integral can be written as: $$ \int_0^2\int_{-y}^{y}(x+y-2xy)dx dy $$

(Note that the function is symmetric, so you can change $x $ with $y$)

If you want, you can also write the limits of the region as: $$ -2<x<2 \qquad \mbox{and}\qquad |x|<y<2 $$

In this case the integral becomes:

$$ \int_{-2}^2\int_{|x|}^{2}(x+y-2xy)dy dx $$

But it is a bit more difficult to evaluate because it requires a split of the $x$ interval.

Answer 2

Because the function $f$ is symmetric around $y=x$. Indeed: $$f(-x,y)=f(y,x), \ -2\le x\le 2, -x\le y\le 2; \\ f(x,y)=f(y,x), 0\le x\le 2, x\le y\le 2.$$ So, the volumes under $f$ for the triangles $(-2,2),(0,0),(2,2)$ and $(0,0),(2,-2),(2,2)$ are equal.