The problem that I am working with is:
Find the volume of the solid of revolution formed by rotating the region $R$ bounded by $y = 4+ x^2,\;x=0,\;y=0,\;and\;x=1$ about the line $y=10$
I have the following so far (using the shell method): $$V=\int_a^b2\pi\,r\,h\,dy\\r = 10-y\\c=2\pi(10-y)\\ h=?$$ $$V=2\pi\int_0^1(10-y)hdy$$ I've been searching for an example in my book where there is a horizontal rotation that is not over the x-axis, but I have had no luck.
Would the height just be $\sqrt{y-4}$ or would it be $10-\sqrt{y-4}$ and why?
We will assume that you really want to use the Shell Method. In that case, we will look at a thin strip of width "$dy$" at height $y$, and rotate it about $y=10$.
The complication is that the thin strip has a different shape from $y=0$ to $y=4$ than it does from $y=4$ to $y=5$.
From $y=0$ to $y=4$, the radius is $10-y$, and the "height" of the cylindrical shell is $1$. Thus the volume obtained by rotating that part is $$\int_{y=0}^4 2\pi(10-y)(1)\,dy.$$ From $y=4$ to $y=5$, the radius is still $10-y$, but the "height" of the cylindrical shell is $1-x$, that is, $1-\sqrt{y-4}$. The required integral is $$\int_{y=4}^5 2\pi(10-y)(1-\sqrt{y-4})\,dy.$$ Calculate the two integrals, and add.
Remark: For this problem, the Method of Washers is less work. Take a cross-section "at" $x$ perpendicular to the $x$-axis. The outer radius is $10$, and the inner radius is $10-y$, that is, $10-(4+x^2)$. The volume is $$\int_{x=0}^1 \pi(10^2-(6-x^2)^2)\,dx.$$