Find the volume of the solid formed by revolving the region bounded by the graphs of
$y=\frac{1}{2}(x-2)^2$ and $y=2$ about the y-axis.
So far, I know the $y-$limits of integration are $y=0$ and $y=2$ and that the volume of the outer donut is $$\pi\int_0^2 \left(2+\sqrt{2y}\right)^2 dy$$ but there is still a hollow cone inside of which I do not know how to calculate the volume in such a way that I could use the washer method.
On
Because $y = \frac{1}{2} (x-2)^2$ is not a one-to-one function, you do not get a single function back when you have to invert it in order to get the boundaries of your "washers". The inversion gives $x = 2 \pm \sqrt{2y}$, so you have correctly found the "outer radius" to be $x = 2 + \sqrt{2y}$, which is a function representing the "upper half" of the "horizontal" parabola, as viewed from the y-axis. The "lower half" of the parabola, $x = 2 - \sqrt{2y}$ is the other function from the inversion; this will serve as the "inner radius" for your slices.
So your integral will be $\pi \int^{2}_{0} [2 + \sqrt{2y}]^2 - [2 - \sqrt{2y}]^2 dy$. This actually simplifies a fair bit when you multiply out the binomials.
Solving for $x$ in terms of $y$, we get $x=2\pm\sqrt{2y}$. For the $y$ in our range, the radius of cross-section of the "hole" at height $y$ is $2-\sqrt{2y}$. This is the inner radius of our washer. The outer radius is the $2+\sqrt{2y}$ of your post.
Thus our volume is $$\pi\int_0^2 \left[(2+\sqrt{2y})^2-(2-\sqrt{2y})^2\right]\,dy.$$ It is useful to simplify the integrand before integrating. There is a whole lot of cancellation going on.
Alternately, we can use cylindrical shells. Often, for rotation about the $y$-axis of a region bounded by $y=f(x)$, that leads to simpler calculations. Not this time. But for the sake of comparison, note that cylindrical shells lead to the integral $$\int_0^4 (2\pi x)\left(2-\frac{1}{2}(x-2)^2\right)\,dx.$$