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Here's a question from one of my exercises,

Exercise 14. Let $C$ be a curve in $\mathbb{R}^2$ given by parametric equations $x=f(t)$, $y=g(t)$. Let $S$ be the surface of revolution of the curve $C$ about the $y$-axis (something like the one shown in Figure 1).

(a) Parametrize the surface $S$. Parameters are $t$ and $\theta$, where $0\leq\theta\leq2\pi$.

(b) Apply part (a) to parametrize the torus of revolution (see Figure 2) obtained by rotating the circle of radius $b$ centered at $(a,0)$ about the $y$-axis. (Assume $a>b$ here.)

I've got part (a) and (b), but how should I go about finding the surface area of this torus?

Here's my solution to first 2 parts:

(a) Applying cylindrical coordinates, since rotation is about $y$-axis,

let $x= r \cos \theta$, $y = y$ and $z = r \sin \theta$, at $\theta=0$ we get $f(t) = r$, hence the parameterization is:

$$x = f(t) \cos \theta\quad y = g(t) \quad z = f(t) \sin \theta$$

for $a \leq t \leq b$ and $0 \leq\theta\leq 2\pi$.

(b) In the plane $x,y$ the circle is $x = a + b \cos \psi$, $y = b \sin \psi$. From part (a), we get the following parametrization for the torus:

$$x = (a + b \cos \psi)\cos \theta, \quad y = b \sin \psi, \quad z = (a + b \cos \psi)\sin \theta$$

where $0\leq\psi,\theta\leq 2\pi$.

What is the surface area of this torus?

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It follows from your calculation that we have a map $F:[0,2\pi]\times[0,2\pi]\rightarrow\mathbb{R}^3$, where $$F(\theta,\psi)=((a + b \cos \psi)\cos \theta, b \sin \psi, (a + b \cos \psi)\sin \theta)$$ and the image under $F$ is the torus. Then we have $$F_\theta=(-(a + b \cos \psi)\sin \theta, 0, (a + b \cos \psi)\cos \theta),$$ $$F_\psi=(-b \sin \psi\cos \theta, b \cos \psi, -b \sin \psi\sin \theta).$$ Hence, $$F_\theta\times F_\psi=(-b (a + b \cos \psi)\cos \theta\cos \psi,-b(a + b \cos \psi)\sin \psi, -b(a + b \cos \psi)\sin \theta\cos\psi)$$ which implies that $$\|F_\theta\times F_\psi\|=b (a + b \cos \psi).$$ It's a standard fact from Calculus that the surface area is given by $$\int_0^{2\pi}\int_0^{2\pi}\|F_\theta\times F_\psi\|d\theta d\psi.$$ (see here: Parametric surface) Therefore the surface area of the torus is equal to $$\int_0^{2\pi}\int_0^{2\pi}b (a + b \cos \psi)d\theta d\psi=4ab\pi^2.$$

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  • $\begingroup$ hey, thanks for the great answer! clear and easy to understand. $\endgroup$ – adsisco Nov 18 '11 at 12:26

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