I'm trying to find the volume of the solid obtained by rotating the region between the curves $y=2x$ and $y=x^2$ around the line y=-1 . This is what the graph looks like
I'm mainly struggling due to the fact that I'm rotating over y=-1, and I'm not fully sure how to account for that.
I've basically done: $$\int_{0}^{2}\pi\left( 2x+1 \right)^{2}dx -\int_{0}^{2}\pi\left( 1+x^{2} \right)^{2}dx$$
but I feel like that's wrong? (I added the plus ones since the the radii are the space between the equation and the x-axis plus the space from the y=-1? I have no idea if that's the correct way to do it).
tl;dr: Your work is correct; I just want to give an explanation that might help you be more confident in why.
The way I like to think about these problems is in terms of integrating a cross-sectional area (which is nice because it generalizes to other methods of rotation other than a pure circle, among other reasons).
If we know the cross-sectional area $A(x)$ of a shape in terms of a single variable $x$, with bounds given by $a,b$, then the volume of the rotated shape is given by $$ V = \int_a^b A(x) \, \mathrm{d}x $$ Try to see how this might align for the formulas you've learned. For instance, in a shape that gets no "holes" in the body generated by the rotation, you will have $A(x) = \pi f(x)^2$.
An example might be with finding the area between $y=x$ and $y=x^2-2x$, rotated about $y=4$. Graphically, we have this setup:
The yellow ring represents the cross section. We can find its area fairly easily, since the area of this ring is given by $$ \text{area of ring} = \pi \text{(outer radius)}^2 - \pi \text{(inner radius)}^2 $$ Try to convince yourself that the inner radius for this circle is $4-x$ (you want to think about that white/blank "inner disc") and that the outer radius is $4-(x^2-2x)$ (the yellow disc as if there were no hole). It may additionally have to note that $y=4$ is the vertical coordinate of the center of all of these circles.
From here, once you have the area of the ring, you would just integrate said area over the bounds for $x$.
(This particular example and others can be found on Paul's Online Math Notes.)
In your case, if you envision rotating your graph in your mind, and look at the cross-sections, they will look something like this if viewed front-onward as if the line $y=-1$ is going through your face:
Or, looked at side-on and poorly drawn:
So, assuming we took that cross-section at the distance of $x$, one deduces that the inner radius is $1+x^2$. (You have to go from the line $y=-1$ up one unit, and then $x^2$ more, to reach the parabola.) Likewise, the outer radius is $1+2x$.
Consequently, your formulation is correct.