1
$\begingroup$

A torus $T_{a,b}\subset \mathbb{R}^3$ is described with torus coordinates $$x=(a+\rho \cos\theta)\cos\phi,\space y=(a+\rho \cos\theta)\sin\phi,\space z=\rho\sin\phi,$$ with $a>0$. Now $T_{a,b}$ is described in these coordinates by $$T_{a,b}=\{(\rho,\theta,\phi):0\leq\rho\leq b<a, 0\leq\theta\leq 2\pi,0\leq\phi\leq 2\pi\}.$$ How do I find the volume of this torus? I have no idea what I'm supposed to do here. Thanks in advance.

$\endgroup$
4
  • 1
    $\begingroup$ See this question. $\endgroup$
    – TonyK
    May 4 '15 at 19:28
  • $\begingroup$ But that question isn't my function $\endgroup$
    – user235238
    May 4 '15 at 19:36
  • $\begingroup$ You can use Pappus's theorem: en.wikipedia.org/wiki/Pappus%27s_centroid_theorem $\endgroup$
    – Chappers
    May 4 '15 at 20:10
  • $\begingroup$ @TonyStrong: You might have to use your brain a little bit. $\endgroup$
    – TonyK
    May 4 '15 at 20:41
1
$\begingroup$

i have a nice ideia :)

we have $T_{a,b}\subset\mathbb{R}^3$ described by $x=(a+\rho\cos\theta)\cos\phi,y=(a+\rho\cos\theta)\sin\phi,z=\rho\sin\theta$ with $a>0$ and $T_{a,b}=\{(\rho,\theta,\phi):0\le\rho\le b<a,0\le\theta\le2\pi,0\le\phi\le2\pi\}$ wich a picture is given below.

an illustration of a toroid

lets take $\vec{F}=x\vec{a}_x$, then we have that the divergence of this vector field is

$$\begin{align} \vec{\nabla}\cdot\vec{F}&=\frac{\partial(x)}{\partial x}+\frac{\partial(0)}{\partial y}+\frac{\partial(0)}{\partial z}\\ &=1 \end{align}$$

lets $S_{a,b}=\partial T_{a,b}$ the edge of region $T_{a,b}$(which is the surface of the toroid), by divergence theorem

$$\begin{align} \iint\limits_{S_{a,b}}\vec{F}\cdot d\vec{S}&=\iiint\limits_{T_{a,b}}\vec{\nabla}\cdot\vec{F}dv\\ &=\iiint\limits_{T_{a,b}}dv \end{align}$$

parameterizing the toroid surface $S_{a,b}$ by

$$\begin{align} \vec{r}(\phi,\theta)&=(a+b\cos\theta)\cos\phi\vec{a}_x+(a+b\cos\theta)\sin\phi\vec{a}_y+b\sin\theta\vec{a}_z\\ \vec{r}_\phi(\phi,\theta)&=-(a+b\cos\theta)\sin\phi\vec{a}_x+(a+b\cos\theta)\cos\phi\vec{a}_y\\ \vec{r}_\theta(\phi,\theta)&=-b\sin\theta\cos\phi\vec{a}_x-b\sin\theta\sin\phi\vec{a}_y+b\cos\theta\vec{a}_z\\ \vec{r}_\phi\wedge\vec{r}_\theta&=\alpha(\phi,\theta)\vec{a}_x+\beta(\phi,\theta)\vec{a}_y+\gamma(\phi,\theta)\vec{a}_z\\ \alpha(\phi,\theta)&=b(a+b\cos\theta)\cos\theta\cos\phi\\ \beta(\phi,\theta)&=b(a+b\cos\theta)\cos\theta\sin\phi\\ \gamma(\phi,\theta)&=b(a+b\cos\theta)\sin\theta \end{align}$$

my nice drawn

then the volume of toroids is given by

$$\begin{align} \iiint\limits_{T_{a,b}}dv&=\iint\limits_{S_{a,b}}\vec{F}\cdot d\vec{S}\\ &=\iint\limits_{S_{a,b}}\vec{F}\cdot \vec{n}dS\\ &=\iint\limits_{S_{a,b}}\vec{F}\cdot\frac{\vec{r}_\phi\wedge\vec{r}_\theta}{||\vec{r}_\phi\wedge\vec{r}_\theta||}||\vec{r}_\phi\wedge\vec{r}_\theta||d\phi d\theta\\ &=\iint\limits_{S_{a,b}}\vec{F}\cdot(\vec{r}_\phi\wedge\vec{r}_\theta)d\phi d\theta\\ &=\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\left[(a+b\cos\theta)\cos\phi\vec{a}_x\right]\cdot\left[\alpha(\phi,\theta)\vec{a}_x+\beta(\phi,\theta)\vec{a}_y+\gamma(\phi,\theta)\vec{a}_z\right]d\phi d\theta\\ &=\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\alpha(\phi,\theta)(a+b\cos\theta)\cos\phi d\phi d\theta\\ &=\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}b(a+b\cos\theta)\cos\theta\cos\phi(a+b\cos\theta)\cos\phi d\phi d\theta\\ &=\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}b(a+b\cos\theta)^2\cos\theta\cos^2\phi d\phi d\theta\\ &=\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}b(a^2+2ab\cos\theta+b^2\cos^2\theta)\cos\theta\cos^2\phi d\phi d\theta\\ &=a^2b\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\cos\theta\cos^2\phi d\phi d\theta+2ab^2\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\cos^2\theta\cos^2\phi d\phi d\theta+b^3\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\cos^3\theta\cos^2\phi d\phi d\theta\\ &=a^2b\int\limits_{0}^{2\pi}\cos^2\phi d\phi\int\limits_{0}^{2\pi}\cos\theta d\theta+2ab^2\int\limits_{0}^{2\pi}\cos^2\phi d\phi\int\limits_{0}^{2\pi}\cos^2\theta d\theta+b^3\int\limits_{0}^{2\pi}\cos^2\phi d\phi\int\limits_{0}^{2\pi}\cos^3\theta d\theta\\ &=a^2b\pi0+2ab^2\pi\pi+b^3\pi0\\ &=2\pi^2ab^2\equiv(2\pi a)(\pi b^2) \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy