There's still some room for interpretation in the question, as the relative orientation of the tubes and the inclination hasn't been specified. I'll assume that the cylinder is inclined by rotating it about a horizontal axis that is perpendicular both to the central axis and to the tubes bisecting the circular ends, which are originally vertical (but cannot remain so upon inclination).
Then the liquid takes the form of a cylindrical wedge. By equations (5) and (9) of the linked page, its volume is given by
$$V=R^3\frac hb\left(\sin\phi-\phi\cos\phi-\frac13\sin^3\phi\right)\;,$$
where $R$ is the radius of the circular end, $b$ is the height indicated on the tube, $h$ is the height of the wedge along the axis of the cylinder and $\phi=\arccos\left(1-b/R\right)$. The ratio $h/b$ is the cotangent of the angle of inclination $\alpha$, so
$$V=R^3\cot\alpha\left(\sin\phi-\phi\cos\phi-\frac13\sin^3\phi\right)\;.$$
Substituting $\phi=\arccos\left(1-b/R\right)$ leads to an unwieldy expression that doesn't seem to allow any useful simplifications.
Consider this diagram, representing a side view of the martini glass with "tipped" so that the beverage reaches a maximum distance $x := |\overline{OX}|$ and minimum distance $y := |\overline{OY}|$ from cone vertex $O$:
Let $O^{\prime\prime}$ be the midpoint of $\overline{XY}$, so that it is
the center of the represented ellipse; thus, $|\overline{O^{\prime\prime} X}| = |\overline{O^{\prime\prime} Y}| = a$ is the ellipse's "major radius".
Two observations follow quickly:
- With $H$ the foot of the perpendicular from $O$ to $\overline{XY}$, so that $h := |\overline{OH}|$ is an altitude of $\triangle XOY$ (and congruent to the altitude of the beverage cone itself),
$$a h = |\triangle XOY| = \frac{1}{2} x y \sin 2\theta \tag{1}$$
- With "horizontals" $\overline{XX^\prime}$, $\overline{YY^\prime}$, and $\overline{X^{\prime\prime}Y^{\prime\prime}}$ as shown above, we see that $\overline{O^{\prime\prime} X^{\prime\prime}}$ and $\overline{O^{\prime\prime}Y^{\prime\prime}}$ are necessarily midpoint segments of respective triangles $\triangle XYX^\prime$ and $\triangle XYY^\prime$; therefore,
$$|\overline{O^{\prime\prime}X^{\prime\prime}}| = \frac{1}{2} |\overline{XX^\prime}| = x \sin\theta \qquad\text{and, likewise}\qquad |\overline{O^{\prime\prime}Y^{\prime\prime}}| = y\sin\theta \tag{2}$$
Also, we note that $b$, the ellipse's "minor radius", is the length of a segment perpendicular to $\overline{X^{\prime\prime}Y^{\prime\prime}}$ at $O^{\prime\prime}$ that meets a (semi-)circle with diameter $\overline{X^{\prime\prime}Y^{\prime\prime}}$:
This is the classical construction of the geometric mean of two values, and we have
$$b^2 = |\overline{O^{\prime\prime}X^{\prime\prime}}| |\overline{O^{\prime\prime}Y^{\prime\prime}}| = x y \sin^2 \theta \tag{3}$$
Now, since the volume of a cone whose base is an ellipse with radii $a$ and $b$, and whose height is $h$, is given by
$$V = \frac{1}{3}\cdot(\text{area of base})\cdot(\text{height}) = \frac{\pi}{3} ab h \tag{4}$$
we have
$$V = \frac{\pi}{3}\cdot a h \cdot b = \frac{\pi}{3}\cdot \frac{1}{2}x y \sin 2\theta \cdot \sqrt{ x y \sin^2\theta} = (xy)^{3/2}\cdot\frac{\pi}{6}\sin\theta\sin 2\theta \tag{$4^\prime$}$$
That is, for fixed $\theta$, we can say more simply:
$$V^2 \;\;\text{is proportional to}\;\; (xy)^3 \tag{$\star$}$$
This being so ... If a certain volume of beverage in a "tilted" glass reaches extreme distances $x$ and $y$ from the vertex, and the same volume of beverage in a "level" glass reaches distance $z$ (in all directions) from the vertex, then we clearly have
$$x y = z^2 \tag{$\star\star$}$$
The "level" distance is the geometric mean of the "tilted" extremes! I suspect that, given the tidiness of this relation, there should be a shorter route to it.
Edit to add ... As @hypergeometric has made the observation that (for constant volume) the ellipse's minor axis and the side-view area remain constant, I'll mention how that these facts arise from my analysis. Simply combine $(\star)$ with $(3)$ and $(1)$ to get:
- $V$ is proportional to $b^3$.
- $V^2$ is proportional to $|\triangle XOY|^3$.
Then, the constancy of $V$ implies the constancy of the other values.
Best Answer
An alternative way to solve case 1 is to say,
"When the cup is tilted, I can divide it into two parts (see right hand pic above):
The part from the lip of the cup to the other end of the liquid surface,
The cylindrical part from the end of the liquid surface to the bottom of the cup." (shown shaded greenish).
For the second case, the figure at the left applies. (Only it doesn't .... because the stuff in the pink triangle above the horizontal line doesn't stay to the left of the blue/pink dividing line when we tilt, so the rest of my earlier analysis was wrong).