Liquid levels from the base of a right cylindrical drinking glass

Question

Suppose that a right cylindrical drinking glass with radius r, is partially filled with a green liquid. The amount of liquid in the glass is L cubic units.(A) represents the angle between the major axis of the cylinder and the top of the liquid. H1 represents the distance from the base of the glass to the top of the liquid on the left-hand side of the glass. H2 represents the distance from the base of the glass to the top of the liquid on the right-hand side of the glass. When the glass is left sitting on a table, A = 90 degrees, and H1 = H2. Tipping the glass to the right such that no liquid pours out and the base of the glass is still completely covered will make H2 > H1, and < 90 degrees. What is the expression to solve H1 and H2 in terms of the given values for r, L, and A

I have derived the expression for H1 or the lower level height in terms of the H when A = 90. that is H-r/Tan(A). Intuitively, I have a hunch that H2 = H+r/Tan(A). But I want to understand how to derive it.

Answer 1

I made a diagram which I think matches your description.

We can devide the volume of liquid in the tilted glass into 2 regions: the bottom fully-covered region and the top half-covered region. The volume of the bottom region is $\pi r^2 H1$ and the volume of the top region is $\frac{1}{2}\pi r^2 (H2-H1)$. The volume before the tilt must be equal to the volume after the tilt, which gives us: $$L=\pi r^2 H = \pi r^2 H1 + \frac{1}{2}\pi r^2 (H2-H1)$$

which gives $$H = \frac{H2+H1}{2}$$

From the figure we can also see that $$tan(A) = \frac{2 r}{H2-H1}$$

Putting these two equations together we find $$H1 = H - \frac{r}{tan(A)}$$

and $$H2 = H + \frac{r}{tan(A)}$$

So your intuition was right.