I am asked to reverse the order of integration an include a sketch of the region. I made a sketch in desmos for the purpose of this post.
The integral is
$$ \int_{-1}^{1}\int_{y-3}^{y^2} f(x, y) \ dx \ dy \\ D = \{ \ (x, y): \ \ y-3 \le x \le y^2, \ -1 \le y \le 1 \ \} $$
I said the reverse would be:
$$ \int_{-1}^{1}\int_{x+3}^{\sqrt x} f(x, y) \ dy \ dx \\ D = \{ \ (x, y): \ \ x+3 \le y \le \sqrt x, \ -1 \le x \le 1 \ \} $$
For my sketch:
Am I right in saying this pink area is the region of integration?
Moreover, is my reverse correct?
To reverse the integral, you must switch which variable varies from function to function and which variable varies from number to number, however upon inspecting this graph, you can see that there are actually 4 different functions that $y$ varies between:
On the interval $-4\leq{x}\leq-2$, the $y$ variable varies between $y=-1$ and $y=x+3$.
On the interval $-2\leq{x}\leq0$, the $y$ variable varies between $y=-1$ and $y=1$.
Finally, on the interval $0\leq{x}\leq1$, the $y$ variable varies between $y=\sqrt{x}$ and $y=1$, but also between $y=-1$ and $y=\sqrt{x}$.
Putting this all together, the new integral becomes:
$$\int_{-4}^{-2}\int_{-1}^{x+3} f(x,y)\ dy \ dx \ + \int_{-2}^{0}\int_{-1}^{1} f(x,y)\ dy \ dx \ + \int_{0}^{1}\left(\int_{\sqrt{x}}^{1} f(x,y)\ dy +\int_{-1}^{-\sqrt{x}} f(x,y)\ dy\right)\ dx$$