How much does the CM go up?

$V_{hemisphere} = \frac {4\pi}6 r^3 = \frac {2\pi}{3}$

$V_{cylinder} = \pi r^2 h = 2\pi$

This means that

$\frac {V_{hemisphere}}{V_{cylinder}} = \frac {\frac {2\pi}3} {2\pi} = \frac 13$

of the cylinder has been moved.

Since the hemisphere was moved by 2 units, the center of mass has gone up by $2 \cdot \frac 13 = \frac 23$ units, so the answer is $2+3 = \boxed{5}$ .

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I don't know the conditions for this method to work, but I think the body has to have rotational symmetry.

Consider the new body as formed by three pieces: the original cylinder, of $m_c= \pi r^2h \,\rho=2\pi \rho$ (where $\rho$ is the density) , the top hemisphere, of mass $m_{t}=\frac{2}{3}\pi r^3\rho=\frac{2}{3}\pi\rho$ , and a bottom hemisphere of negative mass $m_{b}=-\frac{2}{3}\pi\rho$ . The superposition of this negative mass and the cylinder creates the hollow base.

To focus on the physics of the problem, I leave it up to you to prove that the CM of a hemisphere of radius $r$ lies at a height $h=\frac{3}{8}r$ above the base center.

Referencing all positions to the cylinder's CM, the CM's heights of the top and bottom hemispheres are $h_{t}=1+\frac{3}{8}=\frac{11}{8}$ and $h_{b}=-1+\frac{3}{8}=-\frac{5}{8}$ respectively.

We now calculate the CM of the resulting system:

$\Delta h=\dfrac{m_c \cdot h_c+m_{t} \cdot h_{t}+m_{b} \cdot h_{b}}{ m_c+\cancel{ m_{t} }+\cancel{ m_{b} } }=\dfrac{ 2\pi\rho \cdot 0+\frac{2}{3}\pi\rho \cdot \frac{11}{8}+(-\frac{2}{3}\pi\rho) \cdot (-\frac{5}{8})}{2\pi\rho}=\dfrac{2}{3}$ Therefore, $a=2, b=3$ and $a+b=\boxed{5}$ .

We can find that the center of mass of hemisphere is $\frac 38$ of its radius from its base (see note). Then taking moment about the old center of mass, considering empty space as negative mass and let the density of the solid be $\rho$ , we have:

$\begin{aligned} 2\pi \rho \Delta h & = \frac 23\pi \rho \left(1+\frac 38\right) + \frac 23 \pi \rho \left(1-\frac 38 \right) \\ \implies \Delta h & = \frac 13 \left(\frac {11}8\right) + \frac 13 \left(\frac 58 \right) = \frac 23 \end{aligned}$

Therefore, $a+b = 2+3 = \boxed 5$ .

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Note: Let the center of the hemisphere of radius $R$ and density $\rho$ be the origin $(0,0,0)$ and its base normal to the $x$ -axis. Then its center of mass from the center is given by:

$\begin{aligned} h & = \frac {\int_0^Rxy^2\ dx}{\frac 23 \pi \rho R^3} = \frac 3{2R^3} \int_0^1 x(R^2-x^2) \ dx = \frac 3{2R^3} \left[\frac {R^2x^2}2- \frac {x^4}4\right]_0^R = \frac 38 R\end{aligned}$

It is convenient to compute the new center of mass in cylindrical coordinates: $\frac{1}{2\pi }\int_0^{2\pi}\int_{0}^{1}\int_{\sqrt{1-r^2}}^{2+\sqrt{1-r^2}}zrdzdrd\theta=\frac{5}{3}$ . The shift is $\Delta h =\frac{2}{3}$ and the answer is $\boxed{5}$ .