For a region bounded by: $$y=x+4,\;y=16-x^2;\;around\;y=-5$$ I understand that I will be using the 'washer' method: $$V =\int_a^b\pi r^2h$$ But I'm having a hard time finding $$r^2 \text{ for}= \pi(r_{out}^2 -r_{in}^2)dx$$
Since the axis of rotation is at $y=-5$ and the higest point is at 16...
is r: $$(r_{out}^2 -r_{in}^2) \Rightarrow \bigg(11-(16-x^2)\bigg)^2 - \bigg(11-(16-x^2)\bigg)^2?$$
Your inner curve is going to be $x+4$ and your outer curve is going to be $16-x^2$. So the radii are : $r_{in} = x+4 + 5 = x+9$ and $r_{out} = 16-x^2 + 5 = 21-x^2$.
Note that this comes from the fact that our inner curve is $y=x+4$. So our total inner radius is going to be $y + 5$, which includes both the distance above the $x$-axis and the distance below the $x$-axis. A similar argument explains the our radius.
See that one of the intersections of the curves is at $x=-4$. So that gives us our lower bound of the integral. I will let you attempt to solve for the upper bound.