What, if any, is the formula to calculate the volume of a torus given the circumference of the tube and the outer circumference of the ring?
[Math] Calculate Volume of Torus Given Circumferences
geometryvolume
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The picture is I think clearer now.
You are looking for the volume of the solid of rotation generated by sweeping a quarter circle around the center of the beaker. Part of the problem is easy and part of it is subtler. If we take the distance (outer radius - inner radius) to be unit, the area of the quarter circle is just $\pi/4$.
The volume is simply that area multiplied by the distance it travels, which is $2 \pi R$, and the only question is, what to use for R? Pappus' Theorem says that we calculate that distance using the geometric centroid, whose coordinates are found as in the edit below (the value is the same in both dimensions, by symmetry).
So the coordinates of the centroid of the quarter circle are $\{ \frac{4}{3\pi}, \frac{4}{3\pi} \}$.
The radius of revolution should be: R = r_inner + $4/3\pi$ (r_outer - r_inner).
So finally the formula is, using R' as r_inner and r = r_outer - r_inner:
$V = A*2 \pi R = r^2 \frac{\pi} {4}2\pi (r \frac{4}{3 \pi}+ R') $
Edit: some detail about the centroid.
We have the area A from above. For $M_x,M_y$ we have
$M_y = \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} x dydx$ = 1/3.
$M_x = \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} y dydx$ = 1/3.
$x_{centroid} = M_y/A = (1/3) /( Pi/4) = 4/3\pi$
$y_{centroid} = M_x/A = (1/3) /( Pi/4) = 4/3\pi$
You have a problem of order: The limits on the $z$ integral depend on $r$, and therefore it has to be done where $r$ is defined. In other words, the integral is $$\int_0^{2\pi}\int_2^4\int_{-\sqrt{a^2-(r-c)^2}}^{\sqrt{a^2-(r-c)^2}} rdzdrd\theta$$
And you are correct that $c = 3$ and $a = 1$
But as noted in the comments, this approach does not work well with the cuts. The problem is, the cuts are flat, which leads to complicated limits of integration for $r$ and/or $\theta$, which will undo the advantages of going to cylindrical representation ("polar" coordinates are in the plane - the 3D analogs are either "cylindrical", as you are using here, or "spherical").
Let's look at the cuts. The first is by the plane $y = -3$ (which is automatically parallel to the x-axis and to the z-axis). This plane is tangent to the "center ring" of the torus. The other plane is $x = 1$, which passes through the hole. Note that these two planes meet inside the torus. To set the up as an integral in $x, y, z$, you need to consider that for given values of $x$, there are in general two separate ranges of values for $y$ that need integrated over, and vice versa. The way to handle this is to break the torus up into quadrants and work each quadrant separately. Within each quadrant, there is only one range for $x$ or $y$ to be integrated. We can also do the same for $z$, limiting attention to $z \le 0$ and $z \ge 0$ separately. But in this case, the geometry is symmetric, so we will pick up the same volume on either side. Thus we can calculate only $z \ge 0$, then double it to get the whole volume.
The equation of the torus surface is $$z^2 +(\sqrt{x^2 + y^2} - 3)^2 = 1$$ So we can set up the integration over the entire first quadrant as $$2\int_0^4\int_0^{\sqrt{16 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$
After cutting, for each quadrant we have:
- Quadrant 1: $0 \le x \le 1, 0 \le y$
$$2\int_0^1\int_0^{\sqrt{16 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$
- Quadrant 2: $x \le 0, 0 \le y$
$$2\int_{-4}^0\int_0^{\sqrt{16 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$
- Quadrant 3: $x \le 0, -3 \le y \le 0$
$$2\int_{-4}^0\int_{-3}^{\sqrt{4 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$
- Quadrant 4: $0 \le x \le 1, -3 \le y \le 0$
$$2\int_0^1\int_{-3}^{\sqrt{4 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$
The $z$ integration is easy, but for the $y$ and $x$ integrations, you will need some trigonmetric substitutions.
Best Answer
By Pappus' centroid theorem, the volume of a torus generated by the rotation of a circle with radius $r$ with its centre on a circle with radius $R$ is just given by $2\pi^2 r^2 R$. In our case we have $\color{purple}{l}=2\pi r$ and $\color{green}{L}=2\pi(R+r)$, hence:
$$\boxed{ \color{red}{V} = \frac{1}{4\pi}\color{purple}{l}^2(\color{green}{L}-\color{purple}{l})}$$