I'm having trouble working out what a shape is called, let alone how to calculate its volume.
I'm looking for a way to calculate the volume of a shape derived from a truncated cylinder, as described here: https://mathworld.wolfram.com/CylindricalSegment.html
My cylinder has one of the heights as 0.0, so it is like a wedge.
What I want to do is cut the shape in half along the diameter of the bottom circle, at an angle perpendicular to the 3d shape's plane of symmetry. Like this:
The resultant shape will have two semicircles touching at the apex of their arcs, and a rectangular base
For context, I'm not a mathematician, but a gardener. I'm filling in a section of my yard that is semicircular, with a slope towards the flat edge of the semicircle, and need to know how much material to purchase :-)
The height h1 is 0.1, height h2 is 0.0, and radius is 8
We'll want to introduce coordinates here to make things easier to talk about. Let's put an origin at the centre of the base of the cylinder, make its axis of symmetry the $z$-axis and say the vertical cut is the $yz$-plane. Take the direction of the axes so that the half-wedge you're interested in is in the region $x\ge 0$ and $z\ge 0$.
Say the cylinder has radius $r$ (in this case $r=8$); and for convenience, let's write $h$ for your $h_1$.
The cylinder has equation $x^2+y^2=r^2$ and the oblique cut can be written as $hx+2rz = hr$ (I'm using this form so the wedge narrows to a point at $(r,0,0)$).
Let's consider sections parallel to the vertical cut. As you say in your question, these are rectangles. Each corresponds to a particular $x$ coordinate with $0\le x\le r$.
If we look at these in plan view (ie in the $xy$ plane), we find they have width $2\sqrt{r^2-x^2}$.
Looking at a side elevation (the $xz$ plane), the height is $\frac{h}{2r}(r-x)$.
So the rectangular section $x=X$ has area $$A(X)=\frac{h}{r}(r-X)\sqrt{r^2-X^2}$$
To find the volume, we want to add all these areas up; so we integrate: $$V=\int_0^r \frac{h}{r}(r-X)\sqrt{r^2-X^2} dX$$
This looks worse than it is; I can write out the steps but the way to solve it is to expand the multiplication to get two integrals; one is a square root, the other an inverse sine.
We find $$\boxed{V=\frac{h}{12} (3\pi-4) r^2}$$