A specific example would be revolving the area between $x^2-5$ and $5x$ below the $x$-axis about $y=-2$
PS - in general, I am assuming that revolving about any other horizontal or vertical line entails subtracting the relevant variable from the constant value - is that accurate?
Start with drawing a picture. In this case, think washer or donut instead of disk. But first, think of how you'd calculate the area of a donut. You'd say
$$A = \pi R^2 - \pi r^2$$
where $R$ is the outer and $r$ is the inner radius. This problem is similar, except we're stacking all these washers as we march from $x=0$ to $x=5$ where the two curves intersect.
The outer radius goes from $y=-2$ to the top curve, $5x$ and the inner radius goes to $x^2$. Draw this! You ought to see
$$R = 2 + 5x$$ $$r = 2 + x^2$$
Next, set up your integral
$$\pi \int_{x=0}^5 R^2 - r^2 dx$$
It looks like the problem changed to the curves $x^2 -5$ and $5x$. This is now unclear. Do you want to rotate the area bounded by those two curves and the x axis, $y=0$? If so, the rotated region will overlap itself when rotated about $y=-2$ ...