Poured out water - paraboloidal cup

Define a $z$ -axis along the axis of the cup, and an $xy$ -plane at its base. Then the curved surface of the cup has equation $z=\sqrt{\frac23}\left(x^2+y^2-4 \right)$ (found by substituting the given values) and the plane of the surface of the water (after pouring) has equation $z=\frac{x-4+12\sqrt2}{\sqrt3}$ (assuming the tilting happens in the plane $y=0$ )

The region whose volume we want to find is therefore defined by $z>\sqrt{\frac23}\left(x^2+y^2-4 \right)\;\;\text{and}\;\;z>\frac{x-4+12\sqrt2}{\sqrt3}$

(the points inside the cup and above the water.) It's possible to integrate this numerically but I went for a Monte Carlo simulation.

In the plane $y=0$ , the surface of the water meets the cup at two points: $\left(4,0,4\sqrt6 \right)$ (ie the lip of the cup) and $\left(-4,0,\frac{16 + 25 \sqrt2}{2 \sqrt3} \right)$

The region we're interested in therefore lies within the region $-4< x<4,\;\;-4< y<4,\;\;\frac{16 + 25 \sqrt2}{2 \sqrt3} < z < 4\sqrt6$

By randomly picking points in this region, and testing them with the above inequalities, we can estimate the proportion of the total volume that corresponds to the poured off water; this turns out to be around $\boxed{101.6}$ .

Align the tumbler axis along the $x$ -axis. The parabola is of the form $z^2 = ka$ . Let $x_0$ be the $x$ -coordinate of the bottom of the tumbler. Then

$\begin{cases} 2^2 = kx_0 \\ 4^2 = k(x_0 + \sqrt{96}) \end{cases} \implies k = \dfrac 3{\sqrt 6}$ and the equation of the parabola is $z^2 = \dfrac 3{\sqrt 6}x$ .

The water surface is an elipse. Align its major semiaxis on the $xz$ -plane. Then an elementary slice $\delta x$ of the water poured out has a cross-section of a circle segment of radius $r$ center at the $x$ -axis and its base $h$ from the center. Then the cross-sectional area $A = r^2 \cos^{-1}\left(\dfrac hr \right) - h\sqrt{r^2-h^2}$ and the volume of water poured out is given by:

$V = \int_a^b \left(r^2 \cos^{-1}\left(\dfrac hr \right) - h\sqrt{r^2-h^2}\right) dx$

where $a$ and $b$ are the $x$ -coordinates of the near and far ends of major semiaxis for the $z$ -axis. We note that $r$ follows the parabolic curve $r = \sqrt{\dfrac {3x}{\sqrt 6}}$ and $h$ is given by the blue line. We note that $4^2 = \dfrac 3{\sqrt 6}b \implies b = \dfrac {16\sqrt 6}3$ , then the coordinates of the lowest point of the tumbler is $\left(\frac {16\sqrt 6}3, - 4 \right)$ . Then the equation of the blue line is given by

$\begin{aligned} \frac {y+4}{x-\frac {16\sqrt 6}3} & = - \sqrt 3 \\ \implies h & = 16\sqrt 2 - 4 - \sqrt 3 x \end{aligned}$

Since $y^2 = \dfrac {3x}{\sqrt 6}$ , then

$\begin{aligned} (16\sqrt 2 - 4 - \sqrt 3a)^2 & = \frac 3{\sqrt 6}a \\ \implies a & = \frac {65-8\sqrt2 - \sqrt{129-16\sqrt 2}}{2\sqrt 6} \end{aligned}$

With $r(x)$ , $h(x)$ , $a$ , and $b$ know we can find $V$ by solving the integral above. Of course solving it manually is quite formidable. I used a simple Python code below and $V \approx \boxed{101.58}$ .