Is there a way to modify the solid of revolution integral to allow for solids of increasing and decreasing radius?

Question

I am doing a project on tori as they relate to pool floaties and the volume of a normal torus can be calculated by the solids of revolution integral on a circle, Is there a way to modify the integral so that the radius of the circle changes as it is rotated? therefore allowing me to calculate the volume of this shape modified torus where radius of cross section increases and decreases

defined by parametric equations

$F_x (s,t)= (R + r (|(|t/π|-1)|+1)\cos⁡(s)) \cos⁡(t)$

$F_y (s,t)= (R + r (|(|t/π|-1)|+1) \cos⁡(s) ) \sin⁡(t)$

$F_z (s,t)= r (|(|t/π|-1)|+1) \sin(s)$

Also, does Pappus's centroid theorem apply to this shape?

It is my first question and I don't believe there are any repeats of this.

Answer 1

First let's try to simplify where we can.

You can divide your "floaty" in two parts on each side of a vertical plane of symmetry, so that one part is the mirror image of the other. Compute the volume of one part and multiply by $2.$

For one half, it looks like you can let $t$ range from $0$ to $\pi.$ Then $\lvert t/\pi\rvert = t/\pi \leq 1,$ therefore $\lvert t/\pi\rvert - 1 \leq 0,$ therefore $\lvert(\lvert t/\pi\rvert - 1)\rvert = 1 - \lvert t/\pi\rvert = 1 - t/\pi,$ therefore $\lvert(\lvert t/\pi\rvert - 1)\rvert + 1 = 2 - t/\pi.$ That simplifies your parameterization: \begin{align} F_x (s,t) &= (R + (2 - t/\pi) r \cos(s)) \cos(t), \\ F_y (s,t) &= (R + (2 - t/\pi) r \cos(s)) \sin(t), \\ F_z (s,t) &= (2 - t/\pi) r \sin(s). \end{align}

But to represent points inside the floaty, not just on its outer skin, we need a third parameter. Let \begin{align} F_x (s,t,u) &= (R + (2 - t/\pi) ru \cos(s)) \cos(t), \\ F_y (s,t,u) &= (R + (2 - t/\pi) ru \cos(s)) \sin(t), \\ F_z (s,t,u) &= (2 - t/\pi) ru \sin(s), \end{align} that is, $u$ is the relative distance from the circle of radius $R$ inside the floaty, ranging from $0$ on that circle to $1$ on the outer surface of the floaty.

This is a transformation from $\mathbb R^3$ (with coordinates $s,t,u$) to $\mathbb R^3$ (with coordinates $x,y,z$) whose Jacobian is $$ J = - (2 - t/\pi)^2 r^2 R u - (2 - t/\pi)^3 r^3 u^2 \cos(s), $$ with absolute value $$ \lvert J\rvert = (2 - t/\pi)^2 r^2 u \left(R + (2 - t/\pi) r u \cos(s)\right) $$

An intuitive derivation of this is that when we integrate over the three parameters $s,$ $t,$ and $u$ to find the volume, a change of $\mathrm ds$ in $s$ is a distance of $(2 - t/\pi) ru\,\mathrm ds,$ a change of $\mathrm dt$ in $t$ is a distance of $(R + (2 - t/\pi) ru\cos(s))\,\mathrm dt,$ and a change of $\mathrm du$ in $u$ is a distance of $(2 - t/\pi) r\,\mathrm du.$ The product of these three is the "volume element", $\lvert J\rvert \,\mathrm ds \,\mathrm dt \,\mathrm du,$ and you can work out that $(2 - t/\pi) ru \times (R + (2 - t/\pi) ru\cos(s)) \times (2 - t/\pi) r$ equals $\lvert J\rvert$ in the equation above.

For the volume of half the floaty, $s$ goes from $0$ to $2\pi$ (all the way around the circular cross-section of the tube), $t$ goes from $0$ to $\pi$, and $u$ goes from $0$ to $1,$ so we want to integrate $$ \int_0^{2\pi} \int_0^\pi \int_0^1 \left((2 - t/\pi)^2 r^2 u \left(R + (2 - t/\pi) r u \cos(s)\right)\right) \,\mathrm du \,\mathrm dt \,\mathrm ds. $$

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Obviously Pappus's centroid theorem cannot be applied here since the circular cross section keeps chanaging; Pappus assumes you have a uniform cross section throughout the figure. But the form of the integral does suggest something analogous to Pappus's centroid theorem for volume: namely, the cross-sections of the volume perpendicular to the circle $x^2 + y^2 = R^2$ all have centroids on that circle, and we can "straighten out" the circle while keeping all those cross-sections perpendicular to the line through their centroids without changing the volume of the region. We end up transforming one half of the floaty into a circular frustum with bases $2r$ and $r$ and height $\pi R$ (the length of half the circle). The difference between this and Pappus's theorem is that the formula from Pappus's theorem assumes that the "straightened-out" volume is a cylinder, which has a simpler volume formula than the frustum.

Answer 2

This is not an answer to the method you asked for, but I think a general formula for the volume enclosed by a parametric surface might be easier than constructing a surface of revolution approach. Given that Gauss's theorem states for a smooth vector field $\vec{F}$ we have $\newcommand{\oiint}{{\subset\!\supset}\llap{\iint}}$ $$ \iiint_V \nabla\cdot\vec{F} \, \mathrm{d}V = \oiint_S \left(\vec{F}\cdot\hat{n}\right)\, \mathrm{d}S \tag{1} $$ Also, if the surface $S$ is parametrized by a curve $\vec{r}:T\subseteq \mathbb{R}^2 \to\mathbb{R}^3$ given as $\vec{r}(s,t) =(r_x(s,t), r_y(s,t), r_z(s,t))$ then $$ \oiint_S \left(\vec{F}\cdot\hat{n}\right)\, \mathrm{d}S = \iint_{T} \vec{F}\left(\vec{r}(s,t)\right) \cdot\left(\frac{\partial \vec{r}}{\partial s} \times \frac{\partial \vec{r}}{\partial t}\right)\, \mathrm{d}s\,\mathrm{d}t \tag{2} $$ Now, if we choose a vector field $\vec{F}$ such that $\nabla\cdot\vec{F} =1$ then the LHS of $(1)$ is just the volume $V = \iiint_V \mathrm{d}V$. So taking $\vec{F}(x,y,z)=(0,0,z)$ its divergence is indeed $1$, so combining $(1)$ and $(2)$ with this specific vector field we get $$ V = \iint_{T} \left(0,0,r_z(s,t)\right) \cdot\left(\frac{\partial \vec{r}}{\partial s} \times \frac{\partial \vec{r}}{\partial t}\right)\, \mathrm{d}s\,\mathrm{d}t = \boxed{\iint_{T} r_z\left( \frac{\partial r_x}{\partial s}\frac{\partial r_y}{\partial t} -\frac{\partial r_y}{\partial s}\frac{\partial r_x}{\partial t} \right)\, \mathrm{d}s\,\mathrm{d}t} $$ as an integral for the enclosed volume.