Picture of full problem where Dan's share is part of interest:
The image shows the full problem. The issue is raised when calculating Dan's share of the drink.
(1) Equation of plane in Dan's share must pass through $(x,y,z)=(0,0,0)\space and\space (0,R,h)$. Therefore the CARTESIAN equation of the plane is: $$ z=\frac{hy}{R}$$ and the cylindrical equation is: $$z=\frac{h}{R}r\sin\theta$$
(2) The jacobian in cyldindrical coordinates is $r$, therfoe the volume element is: $$dA=r dr d\theta dz$$
(3) The limits of Dan's portion in cylindrical coordinates are: $$z: 0 \rightarrow \frac{h}{R}r\sin\theta \hspace{0.5cm}\vert\hspace{0.5cm} \theta: 0\rightarrow \pi \hspace{0.5cm}\vert\hspace{0.5cm} r:0\rightarrow R $$
(4)The triple integral for Dan's share is therefore:
$$\LARGE \int_{0}^{R}\int_{0}^{\pi}\int_{0}^{\frac{hr\sin\theta}{R}}r\,\, dz d\theta dr$$
(5) Evaluating: $$\large \int_{0}^{R}\int_{0}^{\pi}\frac{h\sin\theta}{R}r^{2}\,\, d\theta dr$$
$$\large \int_{0}^{R}\left[\frac{-h}{R}r^{2}\cos\theta\right]_{0}^{\pi}\,\, dr$$ $$\large \int_{0}^{R}\frac{2h}{R}r^{2}\,\, dr$$
(6) Final answer is therefore : $$\LARGE \frac{2}{3}hR^{2}$$
This seems very odd to me as a semicircular based shaped has a volume that is independent of $\pi$. Is this true or have I made a mistake?
A related question is why does this shape not count as a pyramid? It has a shape that goes to a point. I thought that was the condition for a pyramid. But that would mean the area should be $\frac{1}{6}\pi R^{2}$.
Let us calculate the volume of the body in a more straightforward (but by no means a better) way:
Consider horizontal slices of the body. They are circular segments with the area ($\phi$ is one half of $\theta$ from the wikipedia page): $$ A_\phi =R^2\left(\phi-\sin\phi\cos\phi\right), $$ with $\cos\phi=y/R=z/h$.
Therefore: $$ V=\int_0^h A_{\phi(z)}dz=R^2h\int_0^{\pi/2} \left(\phi-\sin\phi\cos\phi\right)\sin\phi\, d\phi\\=R^2h\left[\sin\phi-\phi\cos\phi-\frac{1}{3}\sin^3\phi\right]_0^{\pi/2}=\frac{2}{3}R^2h. $$ This confirms the correctness of the result.