Unsure of meaning of question A glass of water is rotating about its axis at constant angular velocity $\omega$. Let $y=f(x)$ denote the equation of the curve obtained by cutting the surface of the liquid with a plane passing through its axis of rotation. Show that $f'(x) = \frac{\omega^2}{g}x$, where $g$ is the acceleration of free fall. I don't understand where the plane is or what $f$ represents. I will assume that the glass is upright with its axis of rotation passing through the origin. Is $f$ the height of the water at a cross section that is through the centre of the glass?

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The glass is rotating about that vertical line, and so the water bulges as such. The plane passes through as in this image. $f$ is the equation of that bulge, so yes, the height of the water at the cross section.