$$\iint_{G}\!x^2\,\mathrm{d}x\mathrm{d}y$$
where $G := \left\{(x,y)\in\mathbb{R}^{2}\,;\,|x|+|y| \le 1\right\}$
How does one go about finding the boundaries of these types of integrals? I did fail at searching for examples like this as I don't even know their name(if they do have a specific one).
Oh and also Happy New Year in advance!
On
In double integrals of the form $\int (\int f(x,y)\,dx)\,dy$: the limits of the outer integral are the largest possible values of $y$ over the entire domain of integration; then, for every fixed $y$, the limits of the inner integral are the largest and smallest values of $x$ that can occur in conjunction with that particular value of $y$ (in particular, these are often functions of $y$, whereas the outer limits are always constant).
How big and small can $y$ get in the region $G$?
Given a fixed value of $y$, how big and small can $x$ get (as a function of $y$) so that the point $(x,y)$ is still in $G$?
On
You have to solve the inequalities for the four cases:
$x\geq0\,\land\,y\geq0\,\land x+y\leq1$, the triangle with vertices at $(0,0),(0,1),(1,0)$, limited by the function $y=1-x$
$x\geq0\,\land\,y<0\,\land x-y\leq1$, the triangle with vertices at $(0,0),(0,-1),(1,0)$ with $y=x-1$
$x<0\,\land\,y\geq0\,\land -x+y\leq1$, the triangle with vertices at $(0,0),(0,1),(-1,0)$, with $y=1+x$
$x<0\,\land\,y<0\,\land -x-y\leq1$, the triangle with vertices at $(0,0),(0,-1),(-1,0)$, with $y=-x-1$
So, $x$ runs between $0$ and $1$ or $-1$. The integral is better solved splitting it in two or four for a clearer calculation.
Sketch the region $G$.
It is now clear that $$G=\left\{(x,y)\in\mathbb R^2: |y|\le 1-|x|\,,\,|x|\le 1\right\}$$
So,
\begin{align} \iint_G x^2 \,dx\,dy&=\int_{-1}^{1}x^2\int_{-1+|x|}^{1-|x|}\,dy \,dx \\&=\int_{-1}^1 2x^2 (1-|x|)\,dx \\&=2\int_0^1 2x^2(1-x)\,dx \end{align}