Why do two half-toruses add up to the same volume? We have a small torus $A$ with $R=\frac{11}{2}+0.0005$ and $r=0.0005$. Look at it from the top, and cut along the circle $R$ traces when spun around the center of the torus, and we get the inner and outer half of a torus, inner half clearly with less volume compared to the outer half.
Another similar torus $B$ with $R=\frac{13}{2}$ and $r=0.0005$ is also cut in a similar manner. The outer half clearly has more volume than the inner half.
(I have noted that the different halves are not equal to the volume of half cylinders since one side is flat and thus the volume displacement will not be equal to the original.)
Now, I calculate the volumes of the inner half of A, and the outer half of B using shell integration by setting the center of the torus at $(0,0)$ and rotating the equation of a circle about the line $x=0$, and get $2\cdot\int^{5.5}_{5.4995}\left(2\pi\cdot x\cdot\sqrt{.0005^2-\left(x-\frac{11}{2}\right)^2}\right)\text{ d}x=\frac{\pi(8250\pi-1)}{6000000000}$ and $2\cdot\int^{6.5005}_{6.5}\left(2\pi\cdot x\cdot\sqrt{.0005^2-\left(x-\frac{13}{2}\right)^2}\right)\text{ d}x=\frac{\pi(9750\pi+1)}{6000000000}$ respectively.
I then calculated the same volumes assuming they were equal to half cylinders with the height equal to the circumference of a circle with radius $R$, and got $\frac{11\pi^2}{8000000}$ and $\frac{13\pi^2}{8000000}$ respectively.
I notice something really odd - the volume(assuming half cylinders) for the inner half of A is $\frac{\pi}{6000000000}$ larger than the result I got from shell integration, and the outer half of B was exactly $\frac{\pi}{6000000000}$ smaller.
Why is this? These two toruses clearly have different $R$, and I don't see any reason why the volume inaccuracies for the inner half of $A$ and the outer half of $B$ (when you calculate them as half cylinders) would be the same and cancel each other out if we add them?
Thanks,
Max0815
 A: First, let's define distinct symbols for the "$R$" dimensions of the two toruses:
$R_1 = 5.5,$ $R_2 = 6.5.$
As an attempt at an intuitive understanding,
let's consider first the half cylinder that you are comparing to the "outer half"
of the larger torus.
You used a cylinder of height $2\pi R_2.$
But you cannot just bend this cylinder around a circle of radius $R_2$ to form the "outer half" of the torus, because the very outermost part of the "outer half" has to fill all the circumference of a circle of radius $R_2 + r.$
the particular strip of the torus immediately next to that circle
has to have length $2\pi R_2 + 2\pi r,$ and you only have material of length
$2\pi R_2.$
So let's cut the half cylinder into very thin slices perpendicular to its axis.
Now you can take those slices and arrange them around outside of the circle of radius $R_2.$ But there will be tiny wedge-shaped gaps between the slices.
If we measure the thickness of each wedge at its thickest point,
which is along the circle of radius $R_2 + r,$
the total thickness of the wedges will be $2\pi r.$
Now try to take the other half cylinder and wrap it around the inside of the circle of radius $R_1.$ You cannot do it, because now there is too much material:
along the innermost part of the torus you have room only for material of length
$2\pi R_1 - 2\pi r,$ but you have material of length $2\pi R_1.$
To make the half-cylinder fit in this space, you can slice it into many thin slices
perpendicular to its axis, and then slide an even thinner wedge off each of the thin slices. The total thickness of wedges that you have to slice off will be $2\pi r.$
The wedges you have to slice out of the smaller half-cylinder look a lot like the wedges that are missing from the larger torus: laid on a table, each has the shape of a semicircle of radius $r$; the thin edge of the wedge is along the diameter of the semicircle, and the thickest part of the wedge is halfway along the semicircular arc.
If you cut each half-cylinder into the same (large) number of slices, you have the same number of wedges cut from one half-cylinder as the number of gaps between slices of the other half-cylinder; the wedges are all (just about) identical,
and the total thickness of each set of wedges is the same.
So the wedges you cut off one half-cylinder will (just about) exactly fill all the gaps between slices of the other half-cylinder. (In the limit as the number of slices grows without bound, we can drop the "just about": the wedges fit the gaps exactly.)
The difference in volume between each half-cylinder and the "inner half" or "outer half" of the corresponding torus is just the total volume of wedges you have to cut off or fill in. And since it is the same set of wedges, it is the same volume.
As you can verify, the difference you found in each case between the "half torus" volume and the half-cylinder volume was just exactly the volume of a sphere of radius $r = 0.0005.$

As a bonus, you can assemble the wedges into a sphere by putting the thin edge of each wedge on a shared axis. The circumference of the equatorial circle of the sphere is $2\pi r,$ equal to the total thickness of the wedges
(whose thickest parts are all lined up along that equatorial circle).

More rigorously, in terms of Pappus' Theorem,
the centroid of the semicircle for the "outer half" of the large torus
moves along a circle of radius $R_2 + 4r/(3\pi).$
The volume of that "half" of the torus is therefore
$(R_2 + 4r/(3\pi))\pi r^2,$
exactly $(4r/(3\pi))\pi r^2$ greater than the volume of the corresponding half-cylinder.
The centroid of the semicircle for the "inner half" of the small torus
moves along a circle of radius $R_2 - 4r/(3\pi).$
The volume of that "half" of the torus is therefore
$(R_2 - 4r/(3\pi))\pi r^2,$
exactly $(4r/(3\pi))\pi r^2$ less than the volume of the corresponding half-cylinder.
The volumes added and subtracted are equal because the centroid of one semicircle is the same distance outside the circle of radius $R_2$ as the centroid of the other semicircle is inside the circle of radius $R_1.$
And when you add or subtract equal amounts from the radiuses of any two circles, you add or subtract equal amounts from their circumferences.
A: For many problems it is advantageous to not to plug in numbers too early.  If you can get a generic formula, then it's easier to analyze the working of the problem, and it's also easier to check that the units / dimensions are right.  And in general it's less tedious to write one symbol like $r$ than to carry around strings like "0.0005 cm" and such.
Additionally, in the case of a torus there is Guldin's Rule which states that the volume of a body of revolution is
$$\text{Area-that-revolves} \times \text{Length-of-path-tracked-by-center-of-mass}^1$$

*

*Each half-torus is generated by rotating a half-disc of radius $r$ and area $A=\pi r^2/2$.

*The centroid of a semicircle of radius $r$ lies at $r_c=\dfrac4{3\pi}r$ away from the flat side and on the axis of symmetry. I looked the value up, but it's straight forward enough to calculate on your own$^2$.

*The paths tracked by the centroids is thus $\ell_\pm=2\pi(R\pm r_c)$ where $\ell_+$ is for the outer half-torus and $\ell_-$ for the inner one.

Guldin's rule then yields the volumina
$$\begin{align}
V_\pm
&= \ell_\pm\cdot A
= 2\pi(R\pm r_c) \cdot \frac12 \pi r^2
= \pi^2 r^2 \left(R \pm \frac4{3\pi}r\right) \\
&= \pi^2 r^2 R \,\pm\, \frac43 \pi r^3 
\end{align}$$
which we can express as
$$ V_\pm = \frac12 V_\text Z \,\pm\, \frac43 \pi r^3 $$
where
$$V_\text Z = 2\pi R\cdot\pi r^2$$
is the volume of a cylinder of radius $r$ and length $2\pi R$.  (And the summand after the ± is (incidentally?)$^3$ the volume of a sphere of radius $r$.)
Now we can plug in numbers, where I won't do the $V_\text Z$ part.  Your $r=0.0005=5\cdot10^{-4}$ gives
$$\begin{align}
V_\pm - \frac12 V_\text Z
&=  \frac43 \pi r^3 \\
&= \frac43 \pi \cdot 0.0005^3 
= \frac43 \pi \cdot 1.25 \cdot 10^{-10} \\
&= \frac \pi{6 \cdot 10^9}
\end{align}$$
And there you have it!

Why is this?

Because the difference in volume is independent of $R$ and only depends on $r$.  This is much easier to infer from formulae than by staring at magic numbers :-)  ...and there wasn't even a need to plug in $r=0.0005$.

$^1$Perhaps "Centroid" is more common in English than center of mass.
$^2$Tthe needed integral is hidden in your calculation already.
$^3$Actually not an incident; it's nice to see how the sphere is being "generated" here.
