I have a shape $S$ of area $A$, but I do not know what $A$ is. However, I do have the coordinates of $N$ points (the coords are $(x_i,y_i)$ for $i\in[1,N]$) that follow a sampling distribution $D$. The probability density of $D$ is $0$ on points not enclosed by $S$ and uniform for points enclosed by $S$.
What is the estimation of $A$?
The area of any shape can be written: $A=\sigma^2\gamma(S)$ where $\gamma$ is a scale factor and $\sigma$ is the 2D radius of gyration. $$\hat A=s^2\gamma(S)$$
Where $s$ is the sample standard deviation.
If we don't know what kind of shape $S$ we have, a simple choice is to treat the underlying area as ellipsoid. We calculate the covariance matrix $$\Sigma=\begin{bmatrix} \operatorname{var}(X) & \operatorname{cov}(X,Y)\\ \operatorname{cov}(X,Y) & \operatorname{var}(Y)\\ \end{bmatrix}$$ The eigenvectors of this matrix are in the direction of the semi-axes of the ellipse we're after and the eigenvalues are the variances in the respective semi-axes. We only need to look at the latter, and considering we desire (the square root of) their product, we simply require the determinant of the covariance matrix because this equals the product of the eigenvalues. $$A=\pi ab$$ If the points come from the ellipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the variance of $X$ is $\operatorname{var}(X)=\frac{a^2}{4}$, and variance of $Y$ is $\operatorname{var}(Y)=\frac{b^2}{4}$. That is to say that the length of the semi-axis is two standard deviations (i.e. $2\times\frac{a}{2}$). $$\begin{align}\hat A &= \pi(2\sigma_1)(2\sigma_2)\\ &=4\pi\sqrt{\lambda_1}\sqrt{\lambda_2}\\&=4\pi\sqrt{\operatorname{det}(\Sigma)}\end{align}$$ Our result is:
$$\begin{align}\hat A &= 4\pi\sqrt{\operatorname{var}(X)\operatorname{var}(Y)-\left(\operatorname{cov}(X,Y)\right)^2} \end{align}$$
