Let $x\_0,\dots,x\_n$ be a collection of variable points in $\mathbb{R}^2$ and let $c>0$ be a fixed constant. Is there any way I could compute an upper bound of the volume of the region in $\mathbb{R}^{2(n+1)}$ consisting of all points where $$\|x\_0\|+\sum\_{i=0}^{n-1}\|x\_i-x\_{i+1}\|\leq c$$and $$\|x\_i\|\leq 1$$ for each $i$?



The volume of a region arising from planar linkages
Let $V$ be the volume of the configuration space that you describe in the question. I'll record here the two trivial upper bounds for $V$ arising from dropping either the second set of inequalities $\|x\_i\|\leq 1$ or from dropping the first inequality. For $c$ sufficiently large or small, these bounds become tight, as I explain below.


Let $r\_i=\|x\_i-x\_{i-1}\|$ for $i=0$ to $n$ with the convention that $x\_{-1}=0$. Of course, $r\_i\geq0$ for all $i$, so $\vec{r}=(r\_0,\dots,r\_n)\in\mathbb{R}^{n+1}$ must lie in the "nonnegative orthant" of $\mathbb{R}^{n+1}$.


Your first inequality states that:


$\sum\_{i=0}^nr\_i\leq c$.


This implies that $\vec{r}$ lies in an "orthogonal $(n+1)$-simplex" in $\mathbb{R}^{n+1}$ whose $n+2$ vertices consist of the $n+1$ vectors $(c,0,\dots,0)$, $(0,c,0,\dots,0)$, etc. as well as the origin $(0,\dots,0)$. (See e.g. the discussion [here](https://en.wikipedia.org/wiki/Simplex#Corner_of_cube).) Let's call this simplex $S$.


In general the second set of inequalities $\|x\_i\|\leq1$ do not play well with these $r\_i$ variables. 


However, note that 


$$\|x\_0\|+\|x\_1-x\_0\|+\cdots+\|x\_{i}-x\_{i-1}\|$$
$$\geq \|x\_0+x\_1-x\_0+\cdots+x\_i-x\_{i-1}\|=\|x\_i\|,$$ 


by the triangle inequality. We thus can see that 


$$\|x\_0\|\leq\|x\_1\|\leq\cdots\|x\_n\|\leq c.(\*)$$


(At this point perhaps I should remark that the first inequality suggests that your linkage is a polygonal chain joining the origin in $\mathbb{R}^2$ $\rightarrow$ $x\_0$ $\rightarrow$ $x\_1$ $\cdots$ $x\_n$ where the sum of the lengths at most $c$ and the second set of inequalities requires that all points $x\_i$ lie within the unit disk.)


The above chain of inequalities $(\*)$ implies that when $c\leq 1$, we lose nothing by dropping the second set of inequalities. In this case we can parametrize your configuration space by first fixing the $r\_i$ variables so that $\vec{r}$ lies in $S$ and then picking a solution to the set of equations $\|x\_i-x\_{i-1}\|=r\_i$.


Geometrically, your configuration space is then a bundle over $S$ where the fiber over the point $\vec{r}$ is a Cartesian product of $(n+1)$ circles with radii $r\_0,\dots,r\_n$ (i.e. the set of solutions to $\|x\_i-x\_{i-1}\|=r\_i$). This has volume $(2\pi)^{n+1}\prod\_{i=0}^nr\_i$. Thus the total volume of the configuration space when $c\leq1$ is the following integral


$$I=(2\pi)^{n+1}\int\_{S}\prod\_{i=0}^n dr\_ir\_i.$$ 


We have $I=V$ for $c<1$ and $I\geq V$ in general. See the note below the line for a potential approach to computing $I$.


We can easily bound $I$ to get:


$$V\leq(2\pi)^{n+1}c^{2n+2}/(n+1)!,$$ 


since $c^{n+1}/(n+1)!$ is the volume of $S$ and $r\_i\leq c$ for all $i$.


For $c$ sufficiently large one can do better with the even easier upper bound 


$$V\leq \pi^{n+1},$$ 


which one gets by dropping your *first* inequality.


In fact, this silly bound becomes tight for $c\geq(2n+1)$. The right hand side is the maximum sum of the distances between the adjacent $x\_i$ when they are constrained to lie within the unit disk of $\mathbb{R}^2$ -- place $x\_{2i}$ at $(1,0)$ and $x\_{2i-1}$ at $(-1,0)$ for $i\leq n/2$ so that $\|x\_0\|=1$ and $\|x\_i-x\_{i-1}\|=2$ for $i=1$ to $n$. 




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Aside:


It may be possible to evaluate the integral $I$ more precisely, using techniques described in [this nice answer of Igor Rivin to another MO question](https://mathoverflow.net/a/202821). See part II-1 of [the notes by Baldoni, Berline and Vergne that he referenced](http://www.cmls.polytechnique.fr/perso/berline/Integral_over_simplex_users-guide.pdf).


I made an attempt below but as you can see, I did not get very far.


Explicitly, let us write for the vertices of $S$, $\vec{s}\_{n+1}=(0,\dots,0)\in\mathbb{R}^{n+1}$ and $\vec{s}\_i=(0,\dots,c,\dots,0)\in\mathbb{R}^{n+1}$ (with the $c$ in the $i+1$th coordinate) for $i=0$ to $n$. Note that the volume of $S$ is $\frac{c^{n+1}}{(n+1)!}$ as it is a corner of the hypercube $[0,c]^{n+1}$. 


The key idea is that $\int\_S r\_0\cdots r\_n dV$ (where $dV=\prod\_{i=0}^n$ is the volume measure in $\mathbb{R}^{n+1}$) is equal to the coefficient of $\xi\_0\cdots\xi\_n$ in 


$\int\_S \exp\left(\langle\vec{\xi},\vec{r}\rangle\right) dV$.


(Where $\vec{\xi}=(\xi\_0,\dots,\xi\_n)$ and $\vec{r}=(r\_0,\dots,r\_n)$ lie in $\mathbb{R}^{n+1}$).


Then Brion's formula gives that 
$\int\_S \exp\left(\langle\vec{\xi},\vec{r}\rangle\right) dV=(-1)^{n+1}(n+1)!(vol(S))\sum\_{i=0}^{n+1}\frac{\exp(\langle\vec{\xi},\vec{s\_i}\rangle)}{\prod\_{j\neq i}\langle\vec{\xi},\vec{s\_j}-\vec{s\_i}\rangle}$.


The right hand side simplifies to


$(-c)^{n+1}\left(\sum\_{i=0}^n\frac{\exp(c\xi\_i)}{c^{n+1}\prod\_{j\neq i}(\xi\_j-\xi\_i)}+\frac{1}{c^{n+1}\prod\_{i=0}^n\xi\_i}\right)=\left(\sum\_{i=0}^n\frac{\exp(c\xi\_i)}{\xi\_i^{n+1}\prod\_{j\neq i}(1-\xi\_j/\xi\_i)}+\frac{(-1)^{n+1}}{\prod\_{i=0}^n\xi\_i}\right)$.


Only the sum from $i=0$ to $n$ can contribute to the coefficient of $\xi\_0\cdots\xi\_n$. However, I was not immediately able to compute the relevant coefficient of $\frac{\exp(c\xi\_i)}{\xi\_i^{n+1}\prod\_{j\neq i}(1-\xi\_j/\xi\_i)}$ in closed form.


