Given a curve $\gamma:\mathbb R\to\mathbb R^2,\ s\mapsto \vec\gamma(s)$ of thickness+ $t$, what is the area covered by it for $s\in[a,b]$?
As a simple example, take a circle of radius $r\ge t$, $\vec\gamma(s) = (r\cos\tfrac sr, r\sin\tfrac sr)$, which therefore results in a ring between radii $r-t/2$ and $r+t/2$, i.e. the area is simply $(b-a)\left((1+t/2)^2-(1-t/2)^2\right) = (b-a)t$.
My approach so far:
The tangent unit vector is given as
$$\vec T(s) = \vec\gamma'(s)$$
(where ' denotes derivation as usual, if $s$ is actually parametrizing by arc length such that $|\vec\gamma'|\equiv1$), and the normal unit vector can be obtained using the first of the Frenet-Serret formulas as
$$\vec N = \frac1\kappa\vec T'$$
with curvature $\kappa = |\vec T'|$. For every $s$ we therefore get the line
$$\vec l(s,u) = \vec\gamma(s) + u\cdot\vec N(s), \quad u\in\left[-\tfrac t2, \tfrac t2\right]$$
The sought area is then obtained by integrating $|\vec l(s,u)|$ weighted somehow with probably the curvature, but that's the part where I'm currently stuck. And things get worse for changes in curvature direction (where $\kappa=0$), and I don't have an idea how to properly treat overlapping areas and loops etc…
+ I'm not sure if the "intuitive" definition of this is enough, I mean for every $s$ the normal at $\vec\gamma(s)$ is extended by $\pm t/2$
Best Answer
$\def\ed{\stackrel{\text{def}}{=}}\\ \def\vg{\vec{\gamma}}\\ \def\vl{\vec{l}}\\ \def\vn{\vec{N}}\\ \def\vt{\vec{T}}$ If you parameterise the curve $\ \vec{\gamma}\ $ with its arc length $\ s\ ,$ and the condition \begin{align} &\vg\big(s_1\big)+u_1\vn\big(s_1\big)=\vg\big(s_2\big)+u_2\vn\big(s_2\big),\\ &s_1,s_2\in[a,b],u_1,u_2\in{\scriptsize\left[{-}\frac{t}{2},\frac{t}{2}\right]}\\ \iff&s_1=s_2, u_1=u_2\label{eq1}\tag{c1} \end{align} (i.e. no part of the thickened curve ever intersects any other part) is satisfied, then the formula for the area of the closed subset $$ A\ed\left\{\vg(s)+u\vn(s)\,\left|\,a\le s\le b,{\small{-}\frac{t}{2}}\le u\le{\small\frac{t}{2}}\right.\right\} $$ of $\ \mathbb{R}^2\ $ enclosed within the lines \begin{align} &\big\{\vg(a)+u\vn(a)\,\big|\ u\in{\scriptstyle\left[{-}\frac{t}{2},\frac{t}{2}\right]}\big\}\ \ \ \text{ and}\\ &\big\{\vg(b)+u\vn(b)\,\big|\ u\in{\scriptstyle\left[{-}\frac{t}{2},\frac{t}{2}\right]}\big\}\ , \end{align} and the curves \begin{align} &\left\{\vg(s)-{\small\frac{t}{2}}\vn(s)\,\left|\,s\in[a,b]\right.\right\}\ \ \ \text{ and}\\ &\left\{\vg(s)+{\small\frac{t}{2}}\vn(s)\,\left|\,s\in[a,b]\right.\right\} \end{align} is exactly the same as it is for the circle, namely $$ (b-a)t\ . $$ The area of $\ A\ $ is, by definition, $$ \iint_A1\,dxdy\ , $$ which can be evaluated by using the multivariate change of variable formula for the change of variables from $\ x,y\ $ to $\ s,u\ .$ From the condition (\ref{eq1}), it follows that the mapping $\ \vl:[a,b]\times\left[{-}\frac{t}{2},\frac{t}{2}\right]\rightarrow A\ ,$ given by $$ \vl(s,u)\ed\vg(s)+u\vn(s) $$ is a differentiable bijection, so its Jacobian matrix \begin{align} J(s,u)&\ed\pmatrix{\frac{\partial\vl}{\partial s}(s,u)&\frac{\partial\vl}{\partial u}(s,u)}\\ &=\pmatrix{\vg\,'(s)+u\vn\,'(s)&\vn(s)}\\ &=\pmatrix{\big(1-u\kappa(s)\big)\vg\,'(s)&\vn(s)}\\ &=\pmatrix{\big(1-u\kappa(s)\big)\vt(s)&\vn(s)} \end{align} must be non-singular, and its Jacobian determinant $$ \det J(s,u)=(1-u\kappa(s))\det\pmatrix{\vt(s)&\vn(s)} $$ must be non-zero over $\ (a,b)\times\left({-}\frac{t}{2},\frac{t}{2}\right)\ .$ It follows from this that $$ 0\le \kappa(s)\le\frac{2}{t}\label{eq2}\tag{c2} $$ for all $\ s\in(a,b)\ $, since otherwise $\ \left(s,\frac{1}{\kappa(s)}\right)\ $ would be a point in $\ (a,b)\times\left({-}\frac{t}{2},\frac{t}{2}\right)\ $ where $\ \det J\left(s,\frac{1}{\kappa(s)}\right)\ $ vanished.
Now, by the above-mentioned change of variable formula, \begin{align} \iint_A1\,dxdy&=\int_a^b\int_{{-}\frac{t}{2}}^\frac{t}{2}|\det J(s,u)|\,du\,ds\\\ &=\int_a^b\int_{{-}\frac{t}{2}}^\frac{t}{2}\big|1-u\kappa(s)\big|\left|\det\pmatrix{\vt(s)&\vn(s)}\right|\,du\,ds\\ &=\int_a^b\int_{{-}\frac{t}{2}}^\frac{t}{2}\big(1-u\kappa(s)\big)\,du\,ds\\ &=(b-a)t\ , \end{align} where $\ \big|1-u\kappa(s)\big|=1-u\kappa(s)\ $ for $\ u\in\left({-}\frac{t}{2},\frac{t}{2}\right)\ $ by virtue of inequality (\ref{eq2}), and $\ \big|\det\pmatrix{\vt(s)&\vn(s)}\big|=1\ $ because $\ \vt(s)\ $ and $\ \vn(s)\ $ are orthogonal unit vectors.