Let $S$ be the surface generated by the circles of radius $b$, find a parametric expression for $S$ Let $C$ be the curve associated  to a regular, simple path $\theta:[0,l]\rightarrow  \Bbb  R^2  $; also assume that $((x'(s))^2+((y'(s))^2=b^2$ and let $S$ be the surface generated by the circles of radius $b$, orthogonal to, and centered in points of the curve $\rho(s)=(\theta(s),0) $.
I need to obtain a parametrization for $S$. For this, I was suggested to note that the normal vector to the plane that contains the circle is $(y'(s),-x'(s),0)$, so that the vector that passes trough $(x(s),y(s),0)$ and a point in $S$ must be orthogonal to this vector. 
I haven't been able to find a parametrization for $S$, mainly because I don't understand the geometric figure being described above. I'd appreciate any help.
The purpose of this exercise is to then use surface integrals to compute the area of $S$, and then, defining a 'torus' as an special case of the latter construction for a circle centered in the origin with radius $a$, with $a>b$ and calculating its area.
 A: Here's a parameterization: writing
$$
\theta(s) = (x(s), y(s))
$$
I'm going to define
$$
H(s, t) = ( x(s), y(s), 0) + \cos(t) (0, 0, b) + \sin(t) (-y'(s), x'(s), 0)
$$
which we can regard as having the form 
$$
H(s, t) = ( x(s), y(s), 0) + \cos(t) \mathbf v + \sin(t) \mathbf w
$$
where $\mathbf v$ and $\mathbf w$ are orthogonal vectors of length $b$, and they evidently span the plane orthogonal to $(x'(s), y'(s), 0)$, the tangent to the "centerline" curve. 
For fixed $s$, as $t$ varies from $0$ to $2\pi$, this describes a circle of radius $b$ in the plane orthogonal to $(x'(s), y'(s), 0)$, and containing the point $(x(s), y(s), 0)$. 
For $t = 0$, as $s$ varies, $H(s, 0)$ describes a curve parallel to the centerline curve $(x(s), y(s), 0)$, offset by $b$ units in the $z$ direction. 
In the event that the curvature $\kappa(s)$ of the original curve $\theta$ is larger than or equal to $1/b$ in absolute value, this surface will have singularities (i.e., the derivative of the map $H$ will not have rank 2 at every point of its domain). If the absolute curvature of $\theta$ is strictly bounded by $1/b$, however, then $H$ will be a nonsingular parameterization. 
One more small note about this parameterization: the partials of $H$ at any point $(s, t)$ actually end up being perpendicular, so the length of their cross product (which comes up in various integrals) is just the product of their lengths, which is $b^2$. 
