Are both of these answers to the following integral correct:
\begin{equation} \int_0^1\int_0^1dx_1 dx_2\delta(1-x_1-x_2)\overset{?}{=} 1\text{ or }\sqrt{2} \end{equation}
I can "justify" that the integral is equal to $1$ by using the following property of the delta function
\begin{align} \int_0^1\int_0^1dx_1 dx_2\delta(1-x_1-x_2)f(x_1,x_2)&=\int_0^1dx_1f(x_1,1-x_1)\\ &=\int_0^1 dx_1\\ &=1 \end{align} where I used that our function is just $f(x_1,x_2)=1$ in the second equality.
However I can also justify that the integral is equal to $\sqrt{2}$, as if you draw the region in $\mathbb{R}^2$ that the delta function localizes the integrand on, namely the line $x_2=1-x_1$ (shown in orange in the figure) you see that the orange line segment has length $\sqrt{2}$. More precisely, I could argue that the integrand secretly contains $\sqrt{g}=1$ where $g$ is the determinant of the $g_{\mu\nu}=\begin{pmatrix}1 & 0\\ 0 &1\end{pmatrix}$ metric. Then if I pull-back this $\sqrt{g}$ onto the line segment I get a factor of $\sqrt{2}$; i.e. take the parametric curve $(x_1,x_2)=(\tau,1-\tau)$ for $\tau\in [0,1]$ then $g_{\tau\tau}=\frac{d x^{\mu}}{d\tau}\frac{d x^{\nu}}{d\tau}g_{\mu\nu}=2$ so $\sqrt{g_{\tau\tau}}=\sqrt{2}$.
Are both answers correct because one needs to specify whether one is integrating over $\mathbb{R}^2$ with or without a metric?
$x_2=1-x_1$ which has length $\sqrt{2}$" />
Perhaps a change of coordinates $$x_{\pm}~:=~x_1\pm x_2 $$ (with Jacobian $=2$) is instructive:
$$\begin{align}I ~=~&\iint_{\mathbb{R}^2}\!\mathrm{d}x_1~\mathrm{d}x_2~\theta(0\!\leq\! x_1\!\leq\! 1)~ \delta (1\!-\!x_1\!-\!x_2)\cr ~=~&\frac{1}{2}\iint_{\mathbb{R}^2}\!\mathrm{d}x_+~\mathrm{d}x_-~\theta(0\!\leq\! x_+\!+\!x_-\!\leq\! 2)~ \delta (1\!-\!x_+)\cr ~=~&\frac{1}{2}\int_{\mathbb{R}}\!\mathrm{d}x_-~\theta(-1\!\leq\! x_-\!\leq\! 1) \cr ~=~& 1. \end{align} $$ Here $\theta$ denotes the indicator/characteristic function.