Volume of convex polytope

Question

Let the very simple convex polytope $P$ be given by $$ P=\left\{(x,y)\in\mathbb{R}^2:0

Geometrically, this is a diagonal in the unit square and has length $\sqrt{2}$.

I used different software packages to compute the volume of $P$.

Mathematica:

In Wolfram's Mathematica we can define

P=ImplicitRegion[0

and then Integrate[1,{x,y}\[Element] P] will yield $\sqrt{2}$.

LattE integrale:

The polytope $P$ can be described by the matrix inequality $A\mathbf{x}<\mathbf{b}$, where $$ \mathbf{x}=\left(\begin{matrix}x\\y\end{matrix}\right),\quad A=\left(\begin{matrix}-1&0\\1&0\\0&-1\\0&1\end{matrix}\right), \quad \mathbf{b}=\left(\begin{matrix}0\\1\\0\\1\end{matrix}\right), $$ and the additional linear constraint $x+y=1$. This can be specified as the LattE integrale input file

5 3
0 1 0
1 -1 0
0 0 1
1 0 -1
1 -1 -1
linearity 1 5

The call

integrate filename --valuation=volume

gives answer 1.

Naive computation:

Naively one could compute the volume as $$ \int_{P}1 = \int\int{\mathbf{1}_{(x,y)\in P}\,\mathrm{d}x\,\mathrm{d}y}=\int\int{\mathbf{1}_{\left\{0

Question: What is the volume of $P$? Do the two different answers $1$ and $\sqrt{2}$ reflect different notions of "volume" in this context, or merely a computational mistake on my part?

Answer 1

Your original double integral is 0, since you are finding the area of a line :) To get the correct value, look at the curve $y = 1-x$ and getting the arc length would require changing the integrand to $$ \int_{x=0}^{x=1} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = \int_0^1 \sqrt{1 + (-1)^2} dx = \sqrt{2} \int_0^1 dx = \sqrt{2}. $$