Suppose you have a hemispherical bowl that is open at the top. You tilt the bowl by an angle $\theta$ in any direction. Then you fill it with water till the water surface reaches the lowest point on the rim of the bowl. If the radius of the bowl is $R$, what is the volume $V$ of water as a function of $\theta$?
My approach:
Let the center of the bowl's top be at the origin of the coordinate frame. Rotate the bowl about the $x$ axis by $\theta$, then the minimum elevation of the rim's circle is at $(0, R \cos \theta, - rR \sin \theta ) $. So we want to fill the spherical cap between $z = -R$ and $z = - R \sin \theta $ with water. The height of this cap is $h = R (1 - \sin \theta)$ so from the table in this Wikipedia page on Spherical Cap, we get the volume as
$ V = \dfrac{\pi h^2}{3} ( 3 R - h ) $
Substituting $h$,
$ V = \dfrac{\pi}{3} R^3 (1 - \sin \theta)^2 ( 2 + \sin \theta) $
The ratio of this volume to the maximum bowl volume (which is $\dfrac{2 \pi}{3} $ is given by
$ f(\theta) = \dfrac{1}{2} (1 - \sin \theta)^2 (2 + \sin \theta) $
Below is the graph of this function. As expected it is a decreasing function, that reaches $0$ at $\theta = \dfrac{\pi}{2} $
