How to calculate this area in $\mathbb{R}^2$? Write the area $D$ as the union of regions. Then, calculate $$\int\int_Rxy\textrm{d}A.$$
First of all I do not get a lot of parameters because they are not defined explicitly (like what is $A$? what is $R$?).
Here is what I did for the first question:
The area $D$ can be written as:
$$D=A_1\cup A_2\cup A_3\cup A_4\cup A_5.$$
Where: 
$$A_1=\{(x, y)\in\mathbb{R}^2: x\geq-1\}.$$
$$A_2=\{(x, y)\in\mathbb{R}^2: y\geq-1\}.$$
$$A_3=\{(x, y)\in\mathbb{R}^2: x\leq1\}.$$
$$A_4=\{(x, y)\in\mathbb{R}^2: x\leq y^2\}.$$
$$A_5=\{(x, y)\in\mathbb{R}^2: y\leq1+x^2\}.$$
First, for me I see that $D$ is the intersection of these regions and not the union. Am I wrong?

P.S. This is a homework.
 A: You should write
$$
D =\{(x,y): -1<x<1, -1<y<1+x^2
\}
-
\{ (x,y): 0<x<1, -\sqrt{x}<y<\sqrt{x}
\}
$$
A: Hint: the area inside each of the parabolic indentations is
$$
\int_{-1}^1(1-x^2)\,\mathrm{d}x
$$
A: The easiest way to do this is to write $A$ as $$
  A = Q \setminus \bigl(
    \underbrace{\{(x,y) \in Q \mid y > 1+x^2\}}_{:=B_1} \cup 
    \underbrace{\{(x,y)  \in Q \mid x > y^2\}}_{:=B_2} 
  \big)
$$
where $Q = [-1,1]\times[-1,2]$ (i.e. a rectangle), $B_1$ is the missing part at the top and $B_2$ the missing part on the right. Note that, per your picture, $B_1$ and $B_2$ are disjoint.
You can further use that the area of $B_1$ is the same as the area under the curve $f(x) = 1 - x^2$ (why?) and that the area of $B_2$ is twice the area under the curve $g(x) = \sqrt{x}$ (again, why?).
A: It will take a while but to write as unions you need to break things down I.e. for the top left section you can write as $\{-1\leq x\leq 1, 0\leq y\leq 1\}\cup\{-1\leq x\leq 1,1\leq y\leq 1+x^2\}$. And continue this for the other areas.
Btw $R$ refers to the whole region you have written as unions, and A just refers to the fact that you have an area integral.
