# Calculating volume of a bell shaped container

Given $r_{b}$, $h_{b}$ and $h_{t}$, what would be the equation for calculating the volume of used space in a bell shaped container. Example sketch

Example sketch

What I have so far:

The shape can be separated into 2 peaces, with the first one being a cylinder for which the volume calculation is easy as $$V_{b}=\pi *r^2*h_{b}$$ The second peace is in a shape of a spherical segment for which the volume equation goes as follows $$V_{t}=\frac{1}{6}\pi h_{t}(3r_{b}^2+3r_{t}^2+h_{t}^2)$$ where $r_{t}$ is the radius of the topmost circle. Given that we only have the height $h_{t}$, we need to calculate $r_{t}$ using equation $$r_{t}=\sqrt{r_{b}^2-h_{t}^2}$$ If we input that into the previous equation we get $$V_{t}=\frac{1}{6}\pi h_{t}(3r_{b}^2+3\sqrt{r_{b}^2-h_{t}^2}^2+h_{t}^2)$$ Now we can reduce the equation to get $$V_{t}=\frac{1}{6}\pi h_{t}(6r_{b}^2-2h_{t}^2)$$ and if we combine both equations we get $$V_{bt}=\pi r_{b}^2h_{b} + \frac{1}{6}\pi h_{t}(6r_{b}^2-2h_{t}^2)$$ If we reduce it even further we get the final equation $$V_{bt}=\pi (r_{b}^2h_{b}+h_{t}r_{b}^2-\frac{h_{t}^3}{3})$$

Is this correct? I don't know how to verify it without getting a bucket of water and measuring the volume of a model. Is there a better way of calculating the volume?

The volume is therefore the two cylinders of radius $r_b$ and (by pythagoras) $(r_b^2 - h_t^2)^{1/2}$ plus half the sphere of radius $h_t$