Using a double integral to find a volume by revolving a region R around the y-axis. The question asked is as follows:
"Find the volume of the solid of revolution obtained by rotating the area bounded by the curves about the line indicated.
$$y = |x^2 - 1|, x=-2, x=2, y=-1.$$ Rotate about y=-2."
The equation given to me by my professor for finding volumes by rotation using double integrals is $V = 2 \pi \iint_R y dA$, where R is the region of integration. To deal with the absolute value, I split the integral into three, as follows:
$$V = 2\pi\left(\int_{-2}^{-1}\int_{-1}^{x^2-1}ydydx + \int_{-1}^{1}\int_{-1}^{1-x^2}ydydx + \int_{1}^{2}\int_{-1}^{x^2-1}ydydx\right)$$
However, I know the above expression is incorrect as the second integral (the middle one) evaluates to a negative value. The above expression also does not incorporate the fact that it must be revolved around y=-2, and I feel this is also part of why it is incorrect.
Has anyone dealt with these kinds of problems? For the life of me, I cannot seem to get many of these problems types correct. Any help would be greatly, greatly, appreciated.
 A: Why not forget the double integral formulas and just focus on the fact that it is a solid of revolution about $y=-2$?  In this case, the volume element is a pancake, or disk, of volume $dV = \pi (y+2)^2 dx$, so the net volume is
$$\pi \int_{-2}^2 dx \, [|x^2-1|+2]^2  = \pi \int_{-2}^{-1} dx \, (x^2+1)^2 + \pi \int_1^2 dx \, (x^2+1)^2 + \pi \int_{-1}^1 dx \, (3-x^2)^2$$
I take it you can do the rest.
A: First, a comment. Presumably you have drawn a picture. Note the symmetry about the $y$-axis of the region we are rotating. It is definitely more pleasant to go from $x=0$ to $x=2$, and then double the result.
We still have to split the integral, but into two parts only. And we get a lot fewer minus signs.
Secondly, the formula you are using is not quite the appropriate one. It would be correct if we were rotating a region entirely above the $x$-axis about the $x$-axis.  Take a tiny $dx\times dy$ rectangle in your region, with say lower left corner at $(x,y)$. The distance of this rectangle from the line $y=-2$ is roughly $2+y$. (Note this also works for our negative $y$, since $y=-2$ is below the region we are rotating.) so the volume swept out by the little rectangle is approximately $2\pi(2+y)\,dx\,dy$. That is what you need to integrate.  
Remark: Or else you can use the formula you were given. Then we must make the the $x$-axis as the axis of rotation. 
Lift everything up by $2$. We are rotating the region between $y=|x^2-1|+2$ and $y=1$ about the $x$-axis. For simplicity use symmetry. Let $D^+$ be the part of the modified region that is in the first quadrant. Then our volume is $2\iint_{D^+} y\,dx\,dy$. 
