I have a water tank, a horizontally placed cylinder with two ends (one at right, the other at the left), these ends have a spherical cap [not hemispherical] of a sphere whose radius is $R$.
Given $L$, the level of water from ground (as in figure), and $W$, the width (i.e. the distance between the top of the spherical cap to its circular base)... I can find the volume of partially-filled cylinder or the cap (but filled not partially filled). Can anyone help me?
On
There is one relatively straightforward thing to try: the surface area of the water in the tank is the rate of change of the volume as the depth of the water increases.
Write an expression showing the surface area of the water as a function of the depth of the water. Then integrate over the appropriate interval (from no water in the tank up to the given level).
Suggestions: A ball of radius $R$ centered at the origin is described by $$ x^{2} + y^{2} + z^{2} \leq R^{2}. \tag{1} $$ The end cap of your tank can be modeled as the set of points $$ x_{0} := R - W \leq x \leq R. \tag{2} $$ Taking the positive $z$-axis to point "upward" (perpendicularly to the fluid surface), the bottom of the tank lies at height $$ z_{0} := -\sqrt{R^{2} - x_{0}^{2}} = -\sqrt{(2R - W)W}. $$ The fluid in the tank therefore lies in the region $$ z_{0} \leq z \leq z_{0} + L. \tag{3} $$ The proposed strategy is to cut the region into slices $x = \text{constant}$ and calculate the area of a slice, then integrate the areas from $x = x_{0} = R - W$ to $x = R$, see (2). The slice at $x$ is the portion of a disk of radius $\sqrt{R^{2} - x^{2}}$ cut by a chord lying at a distance $|z_{0} + L|$ from the center.
Overall, the calculation looks messy but doable. I haven't worked out the details, however, and I haven't given much thought to other coordinate systems.