Double integrals - when do I need to split the area of integration? I am having real trouble determining when I have to split the area of integration into two or more parts. Here's an example:

So in the above example, it seems natural to me that we should split the area of integration into two parts $D_1$ and $D_2$, so the bounds for $y$ can be two functions (for example, the bounds for $y$ in $D_1$ are $y=-x-2$ and $y=\frac{1}{2}x-\frac{1}{2}$).
But, in this example:

We don't have to split the area into two parts. I don't even know if it's possible, but I tried because I have absolutely no clue when to split the area into two parts and when not to do so.
Any help will be appreciated!
 A: As long as you are able to describe the bounds, you don't have to split the interval.
For example, the first question can be written as $$\int_{x_1}^{x_2}\int_{\max(-x-2, 2x-5)}^{\frac12 x -\frac12} \, f(x,y)\,dydx$$
However, it is typical to split it for the simplicity of computation. We typically split it when the rule change.
A: That depends on how is the region aligned to coordinate axes. That is why many a times we use change of variable in such a way that the transformed region is better aligned to coordinate axes to compute the integral.
Here are some common scenarios you can get into -
i) it requires dividing the region into multiple sub-regions to find area regardless of whether we integrate wrt $dx$ first or $dy$ first
ii) it requires dividing the region into two to find area for one order but can be done in single integral in another order.
iii) Simplest scenario is when it can be evaluated in any order without splitting the region.
Let me take a simple example of (ii). The region is  in first quadrant and is bound between $y \leq x \leq 2y$ and $y = 2$.
If we integrate wrt $dx$ first -

Can you see that for any given $y$, the horizontal strips (parallel to x-axis) are bound between same curves, that are $x = y$ and $x = 2y$?
If we integrate wrt $dy$ first -

Can you see that for certain interval of $x$, the vertical strips (parallel to y-axis) are bound between curves $x = y$ and $x = 2y$ but for rest of the interval of $x$, they are bound between $x=2y$ and $y = 2$?
Now take the first example in your question. You can see that if we integrate wrt $dx$ first, horizontal strips are not bound between same curves for all $y$ and same is true of vertical strips if we go $dy$ first. So it will require us to split the region into two regardless of the order of integral.
In the second example, you can clearly see that any order you take, the horizontal and vertical strips both run between two given curves in the interval and hence any order would need just one integral.
Many a times we can figure this out just based on intersection points but it is always a good idea to draw a rough sketch.
Now in the example that I had for (ii), I will add one more constraint $x + y \geq 2$.

Now can you see that integrating wrt $dx$ first requires two integrals whereas integrating wrt $dy$ first requires three integrals?
Lastly, I will leave you with a simple example of how change of coordinate system helps. Here is an example where in polar coordinates, it is one integral but cartesian requires two. Consider region in first quadrant between $1 \leq x^2 + y^2 \leq 9$. Can you see $dx \ dy$ or $dy \ dx$ require you to split your region into two whereas going radially is simply $1 \leq r \leq 3$ for all $\theta \in [0, \pi/2]$?

