Vector Calculus - Tubes

Vector-Calculus - Extra Credit Problem

• This amazing problem is optional, for extra credit.
• Submit your solution to Gradescope. Due date is Dec. 13, 8:00am.
• Your score on this problem will replace your 3rd lowest HW score if it is higher.
• The problem is challenging… Be warned!
• You are not allowed to collaborate or receive any help from another person or resource besides your Packets, HW, textbook, and your notes generated during this course.
• Please pledge your solution with regard to the previous instruction.
• No hints will be provided besides the Solution Guide (and possible corrections).

Problem statement

Volume of a tube

Show that the volume of any tube $\mathcal{T}$ with circular cross section is always $\pi r^{2}L$, where $r$ is the radius of a cross section, and $L$ is the length of the curve that traces the centers of cross sections.

The solution uses many techniques and results from the course. One quantity, the torsion $\tau (s)$, was not taught in this year’s version of the course; it is discussed in the Solution Guide below.

The surface of the tube may be described using the following parametric formula:

$$\mathbf{r}(s,\theta )=\mathbf{q}(s)+r\cos (\theta )\mathbf{N}(s)+r\sin (\theta )\mathbf{B}(s).$$

In this formula, $\mathbf{q}(s)$ for $s\in [0,L]$ is an arclength parametrization of the centroid curve. Points on this curve follow the centers of the cross sections.

The vectors $\mathbf{N}(s)$ and $\mathbf{B}(s)$ are the principal normal and binormal of $\mathbf{q}(s)$ as defined in Packet 05. So, the parameter $s$ controls how far along the tube you are, and $\theta$ controls where around the circumference of the tube you are.

Solution Guide

Step 0. Strategy. Take the vector field $\mathbf{F}=(x,y,z)$. Then $\nabla \cdot \mathbf{F}=3$, so ${\iiint }_{\mathcal{T}}\nabla \cdot \mathbf{F}dV=3\cdot \text{Volume}$. Therefore, by the divergence theorem, the volume is given by $\frac{1}{3}{\iint }_{\partial \mathcal{T}}\mathbf{F}\cdot d\mathbf{A}$ where $\partial \mathcal{T}$ is the surface of the tube. The plan is to compute this surface integral for the flux of $\mathbf{F}$ exiting the tube by using the given parametrization of the surface of the tube.

It is extremely important to remember in your calculations that $s$ is an arclength parameter.

Step 1. Compute ${\mathbf{r}}_{\theta }$ and ${\mathbf{r}}_s$ and $\mathbf{n}={\mathbf{r}}_{\theta }\times {\mathbf{r}}_s$. You should arrive at:

$$\mathbf{n}=(1-\kappa r\cos (\theta ))r\cos (\theta )\mathbf{N}(s)+(1-\kappa r\cos (\theta ))r\sin (\theta )\mathbf{B}(s).$$

In order to make this calculation, you need to remember that ${\mathbf{q}}^{\prime }(s)=\mathbf{T}(s)$ and ${\mathbf{T}}^{\prime }(s)=\kappa (s)\mathbf{N}(s)$. You also need to use a new fact: namely that ${\mathbf{N}}^{\prime }(s)=\tau (s)\mathbf{B}(s)-\kappa (s)\mathbf{T}(s)$. In this formula, $\tau (s)$, called torsion, is a quantity similar to $\kappa (s)$.

Torsion

The torsion function $\tau (s)$ may be defined by the formula:

$${\mathbf{N}}^{\prime }(s)=\tau (s)\mathbf{B}(s)-\kappa (s)\mathbf{T}(s)$$

using components in the orthonormal basis $(\mathbf{T},\mathbf{N},\mathbf{B})$: we know $\mathbf{T}\cdot {\mathbf{N}}^{\prime }=-\kappa$, (apply the product rule to $(\mathbf{T}\cdot \mathbf{N}{)}^{\prime }$) as well as $\mathbf{N}\cdot {\mathbf{N}}^{\prime }=0$ (why?), and then $\tau$ is just whatever coefficient exists on the remaining $\mathbf{B}$ component in the decomposition of ${\mathbf{N}}^{\prime }$ according to the $(\mathbf{T},\mathbf{N},\mathbf{B})$ basis.

You do not need these details to complete the problem, you only need the formula for $N^{\prime }$. Don’t forget that since $\mathbf{B}=\mathbf{T}\times \mathbf{N}$, you know that, e.g., $\mathbf{T}\times \mathbf{B}=-\mathbf{N}$, and similarly for other cross products. (Use the right-hand rule for each one.)

Step 2. Observe that $\mathbf{F}(\mathbf{r}(s,\theta ))=\mathbf{r}(s,\theta )$. Show that:

$$\mathbf{F}(\mathbf{r}(s,\theta ))\cdot \mathbf{n}(s,\theta )=\mathbf{q}(s)\cdot \mathbf{n}(s,\theta )+r^2(1-\kappa r\cos (\theta )).$$

Step 3. Show that ${\iint }_{\partial \mathcal{T}}{r}^2(1-\kappa r\cos (\theta ))d\theta ds=2\pi r^{2}L$.

Step 4. Expand $\mathbf{q}\cdot \mathbf{n}$:

$$\mathbf{q}(s)\cdot \mathbf{n}(s,\theta )=(1-\kappa r\cos (\theta ))r\cos (\theta )\mathbf{q}(s)\cdot \mathbf{N}(s)+(1-\kappa r\cos (\theta ))r\sin (\theta )\mathbf{q}(s)\cdot \mathbf{B}(s).$$

Notice that the dot products do not depend on $\theta$, so you can already perform the $\theta$ integral at this point, and then do the $s$ integral later. Show that the $\theta$ integral will annihilate all terms except this one:

$$-\kappa r^2{\cos }^2(\theta )\mathbf{q}(s)\cdot \mathbf{N}(s).$$

The $\theta$ integral will convert ${\cos }^2(\theta )$ to $\pi$. So it remains to find the value of $-\int \mathbf{q}(s)\cdot (\kappa \mathbf{N}(s))ds$.

Step 5. Using the product rule, derive the identity:

$$(\mathbf{q}\cdot \mathbf{T}{)}^{\prime }=1+\mathbf{q}\cdot \kappa \mathbf{N}.$$

Step 6. Using the identity of Step 5, show that:

$$-{\int }_0^L\mathbf{q}(s)\cdot (\kappa \mathbf{N}(s))ds=L+\mathbf{q}(0)\cdot \mathbf{T}(0)-\mathbf{q}(L)\cdot \mathbf{T}(L).$$

Step 7. Combine Step 6 with Steps 2-4 to find the flux across the surface of the tube.

Step 8. The boundary $\partial \mathcal{T}$ includes the tube surface as well as the end caps. Show that the flux across the end caps (with appropriate orientations) is:

$$\pi r^2(\mathbf{q}(L)\cdot \mathbf{T}(L)-\mathbf{q}(0)\cdot \mathbf{T}(0)).$$

This calculation requires a few steps of its own. You will want to parametrize the end caps, perhaps using:

$$\mathbf{m}(r,\theta )=\mathbf{q}(L)+r\cos (\theta )\mathbf{N}+r\sin (\theta )\mathbf{B}$$

for the endcap at $s=L$, and a similar one for the endcap at $s=0$. (Check your orientation at $s=0$!) (Notice that $r$ is a variable here, while it was a constant for the tube surface.)

Step 9. Combine Steps 7-8 to obtain the total flux exiting the tube. Your answer should be $3\pi r^{2}L$. Based on Step 0, you are done.

Step 10. Pledge your solution and submit.