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?