How to calculate the height h reached by a liquid contained in a cylinder with a hemisphere at only one end? I know that I have to calculate separately the volume of the cylinder and the hemisphere. The expression for the cylinder, in terms of the height h, I already obtained by calculating the cross-sectional area of the cylinder and then multiplying it by the length of the cylinder. But I don't know how to calculate the cross-sectional area of this hemisphere, since I think it must be the same procedure as before, or similar.
The resulting expression for the volume of the cylinder was:
$$V = (R\frac{arcsin(h-R)}{R}+\frac{h-R}{R} \sqrt{R^2-(h-R)^2} +\frac{\pi}{2})L$$
where R is the radius of the cylinder, h is the height a liquid can reach and L is the length of the cylinder
Let $0 \le k \le 1$, it's about calculating the volume:
$$ V(k) = \iiint\limits_{\Omega_1} 1\,\text{d}x\,\text{d}y\,\text{d}z + \iiint\limits_{\Omega_2} 1\,\text{d}x\,\text{d}y\,\text{d}z, $$
$$ \begin{aligned} & \Omega_1 = \left\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2\le r^2, \, x \le 0, \, -r \le z \le (2\,k-1)\,r\right\}; \\ & \Omega_2 = \left\{(x,y,z) \in \mathbb{R}^3 : y^2+z^2\le r^2, \, 0 \le x \le L, \, -r \le z \le (2\,k-1)\,r\right\}. \\ \end{aligned} $$
So, all that remains is to calculate:
$$ \iiint\limits_{\Omega_1} 1\,\text{d}x\,\text{d}y\,\text{d}z = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \text{d}\theta \int_{-r}^{(2k-1)r} \text{d}z \int_0^{\sqrt{r^2-z^2}} \rho\,\text{d}\rho = (3-2\,k)\,k^2\left(\frac{2}{3}\,\pi\,r^3\right); $$
$$ \begin{aligned} \iiint\limits_{\Omega_2} 1\,\text{d}x\,\text{d}y\,\text{d}z & = \int_0^L \text{d}x \int_{-r}^{(2k-1)r} \text{d}z \int_{-\sqrt{r^2-z^2}}^{\sqrt{r^2-z^2}} \text{d}y \\ & = \frac{2(2\,k-1)\sqrt{k\,(1-k)}+\arccos(1-2\,k)}{\pi}\left(\pi\,r^2L\right); \end{aligned} $$
from which what is desired:
$$V(k) = \color{red}{(3-2\,k)\,k^2}\color{blue}{\left(\frac{2}{3}\,\pi\,r^3\right)} + \color{red}{\frac{2(2\,k-1)\sqrt{k\,(1-k)}+\arccos(1-2\,k)}{\pi}}\color{green}{\left(\pi\,r^2L\right)}. $$