Sketch the region of integration for the integral $\int_{-2}^2 \int_0^{2y} f(x,y)dx\,dy$

Question

I am slightly confused when sketching the region for the double integral $$\int_{-2}^2 \int_0^{2y} f(x,y)dx\,dy$$ When I sketch the region I get a triangle in the top right quadrant, however my textbook says the region is two triangles, one in the top right quadrant, one in the bottom left quadrant. I don't understand how this can be if the inequalities for the regions are $0<x<2y$ and $-2<y<2$.

Answer 1

Note that, as mentioned here, $$ \int_a^bf(x)\,\mathrm{d}x=-\int_b^af(x)\,\mathrm{d}x $$ That is, $x$ is between $0$ and $2y$. This means that when $y\lt0$, the range for $x$ is $2y\lt x\lt0$. Thus, $$ \int_{-2}^2\int_0^{2y}f(x,y)\,\mathrm{d}x\,\mathrm{d}y =\underbrace{\int_0^2\int_0^{2y}f(x,y)\,\mathrm{d}x\,\mathrm{d}y}_\text{triangle above the $x$-axis}-\underbrace{\int_{-2}^0\int_{2y}^0f(x,y)\,\mathrm{d}x\,\mathrm{d}y}_\text{triangle below the $x$-axis} $$

Answer 2

$$\int\limits_{-2}^2 \int\limits_0^{2y} f(x,y)dx\,dy=\underbrace{\int\limits_{-4}^0 \int\limits_{-2}^{\frac{x}{2}} f(x,y)dy\,dx}_\text{left triangle}+\underbrace{\int\limits_{0}^4 \int\limits_{\frac{x}{2}}^{2} f(x,y)dy\,dx}_\text{right triangle}$$