Find area bounded by distance from a curve I am looking to find the area of region given by distance $d$ from a curve $f(t) = (x,y)$. In other words, the area covered by drawing a curve with a thick circular pencil of radius $d$ (thanks mvw)
I approached the problem as follows:


*

*Find two curves (the curve "above" $g(t)$ and the curve "below" $h(t)$)

*To find the points above and below $f(t)$, we find the slope of the perpendicular line $m=-{f'(t)}^{-1}$ and then move distance $d$ along it in both directions to get the points above and below

*I then say that $g(t) = f(t) + \frac{dm}{\|m\|}, h(t)=f(t) - \frac{dm}{\|m\|}$

*Then the area within that distance should be $\int{h(t)}-\int{g(t)}$


As far as I can tell though, there is a mistake somewhere as given the simple case of $f(t)=(0,0)$, I should get an area of $\pi$, but I get nothing of the sort.
I think the issue is in step 2? but I am open to other ways of solving this problem.
 A: If $f : I \to \mathbb{R^2}$ and $d \ge 0$ then the area of interest is
$$
A = \{ u \in \mathbb{R}^2 \mid \exists t \in I: \lVert f(t) - u \rVert \le d \}
$$
My guess is that
$$
\pi d^2 \le \lvert A \rvert \le \pi d^2 + 2 s d 
$$
where $s$ is the length of the straightened curve (stretched to a straight line), which is one extremal case, while the curve wound around a point is the other.
A: Let $\gamma(t)$ be a curve with radius of curvature at least $d$ at all points.
Let $W(t)$ be a line segment of length $2d$ perpendicular to $\gamma$
at $\gamma(t)$. That is, $W(t)$ is like a pair of "whiskers" that sweep
out some area as you move along the curve.
Let $W(t_1)$ be disjoint from $W(t_2)$ whenever
$\gamma(t_1) \neq \gamma(t_2)$. That is, the curve never crosses itself
or "sideswipes" (with its whiskers) an area that was already swept out.
Let $A$ be the region swept out.
Then $\lvert A \rvert = \pi d^2 + 2sd$, where $s$ is the path length
of $\gamma.$
For a less "well-behaved" curve that does not satisfy all the conditions
above, cut one or more pieces from it that do satisfy those conditions,
either doing so in a way that gives you no two pieces whose swept regions
overlap, or being careful to keep any overlapping simple enough to
deal with as explained below.
These are the "simple" pieces. Estimate their total area; this is
$2ds_s - A_s$ where $s_s$ is the combined length of all these pieces
and $A_s$ is the amount you have to deduct due to overlapping.
(Computing $A_s$ is a relatively ad-hoc procedure in general;
but if there is no overlap then $A_s = 0$.)
The remaining pieces of the curve are the "complicated" pieces.
Here you do the best you can to cut these into shapes whose
area you can measure, and add this to the total.
If an approximate answer is good enough, this is a good place to
do some approximating.
A possibly helpful fact in dealing with the "complicated" pieces is
that if you have a curve of length $s_1$ whose direction changes a total of
$\theta$ radians to the right (where $\theta$ can be greater than
$2\pi$; that is, we're counting "turns", not just initial and final direction), and if the curve never curves to the left with radius of
curvature less than $d$ nor does the left-hand part of the
"whisker" re-enter any part of the area it already swept out, 
the region swept out by the left-hand part
of the "whisker" is $s_1d + \frac12 \theta d$.
Finally, you have to add whatever the "end caps" at the two ends of
the curve (if they exist) would contribute.
This is at most $\pi d^2$; it can be less in the case where either of
the "end caps" overlaps the region swept out by the whiskers along some
part of the curve.
