One should change the order of integration of $\int_{-1}^2\int_{-x}^{2-x^2} f(x,y) dy dx$
My idea was to first parameterize the area $A$: $$A = \{ (x,y)\in \mathbb{R^2} | -1 \leq x \leq 2 \land -x \leq y\leq 2-x^2 \}$$
This is what $A$ looks like:
$A$" />I now want to write $A$ in terms of $y$, where $x$ depends on $y$: For sure since $ -1 \leq x \leq 2 \land -x \leq y\leq 2-x^2 $ we can see that $y$ must be bigger than $-2$ and smaller than $2$ so we can write:
$$A = \{ (x,y)\in \mathbb{R^2} | -2 \leq y \leq 2 \land ... \}$$
Now comes the tricky part, I must somehow find a way to express $x$ relative to $y$.
In order to do that I have plotted $A$. It seems that $x$ can be parameterized by something similar to $\sqrt{y...} \leq x \leq y$ by looking at the picture.
My questions are as follows:
- How does the parameterization for $x$ look like?
- Is there a general way of finding these parameterizations? Guessing them seems to work fine for very simple examples but already with this example, it is difficult to guess one.
Thank you so much for your time.
There is a clear shift in behaviour as you move across $y=1$. So it makes sense to split the set as $$ A = \{ -2≤y≤1 , \quad \dots\quad \} \cup \{ 1≤y≤2, \quad \dots \quad \}$$ and now the game is to figure out what goes in the blanks. Drawing with axes, + also a line to show where the cut is:
The curvy bit of the boundary is defined by $$ y = 2-x^2\quad (-1≤x≤2)$$ this of course inverts to $x = \pm\sqrt{2-y}$. So for the second bit which jumps between the two curved lines, the condition should be $-\sqrt{2-y}<x<\sqrt{2-y}$.
The first case is easier; the lower boundary is the linear function $y=-x$. So in summary,
$$ A = \{ -2≤y≤1 , -y ≤ x ≤ \sqrt{2-y} \} \cup \{ 1≤y≤2, -\sqrt{2-y}<x<\sqrt{2-y} \}$$ Verification that this answer is correct: