Poured out water - conical cup

Align the axis of the conical frustrum along the $x$ -axis with the point of water surface farthest from the brim on the $z$ -axis as shown. Then the volume of water poured out is the volume between the internal wall of the frustrum and the water surface.

For a value of $x$ , a slice $\delta x$ of this poured water has a surface cross-section of a segment of a circle of radius $r$ with the base of the segment distance $h$ from the center of the circle or the $x$ -axis. Let $\frac \theta 2 = \cos^{-1} \left( \frac hr \right)$ . Then the cross-sectional area $A = r^2 \cos^{-1} \left(\frac hr \right) - h\sqrt{r^2-h^2}$ and the volume of the poured water is given by:

$V = \int_0^\red l \left(r^2 \cos^{-1} \left(\frac hr\right) - h\sqrt{r^2-h^2}\right) dx$

where $\red l$ is the horizontal distance between the brim and the $z$ -axis (red dash line). We note that the gradient of the slanting side of the frustrum $m = \frac 2{\sqrt{10^2-2^2}} = \frac 1{2\sqrt 6}$ . Consider the diameter of the brim

$\begin{aligned} ml + l \cot 30^\circ & = 8 \\ \implies l & = \frac 8{\frac 1{2\sqrt 6}+\sqrt 3} = \frac {16\sqrt 6}{1+6 \sqrt 2} \end{aligned}$

We note that $r(x)$ is given by the top black line

$\begin{aligned} r(x) & = mx + z_0 & \small \blue{\text{where }z_0 \text{ is the }z\text{-intercept.}} \\ r(0) & = z_0 = 4 - ml \\ \implies r & = m(x - l) + 4 \end{aligned}$

And $h (x)$ is given by the blue line (water surface) $h = 4 - ml - \sqrt 3 x$ . With $r$ , $h$ , and $l$ known we can perform the integration (of course not by hand) to find $V$ . I used a simple Python code and the answer is $\boxed{98.16}$ .