Consider the inertia tensor $\bm{I}_x$ defined by
$$ \bm{I}_x \triangleq \int \bm{x} \cdot \bm{x} \, \mathbb{1} - \bm{x} \otimes \bm{x} \, \mathrm{d} \alpha $$The transformed tensor $\bm{I}_{x+\rho}$ corresponding to a translation by $\bm{\rho}$ is obtained by replacing $\bm{x}$ with $\bm{x} + \bm{\rho}$, furnishing
$$ \bm{I}_{x+\rho} = \bm{I}_x + 2 A \,\bm{\rho}\cdot \bar{\bm{\rho}} \, \mathbb{1} - A (\bar{\bm{\rho}}\otimes\bm{\rho} + \bm{\rho}\otimes\bar{\bm{\rho}}) + A (\bm{\rho}\cdot\bm{\rho}\,\mathbb{1} - \bm{\rho}\otimes\bm{\rho}) $$where the centroid $\bar{\bm{\rho}}$ and area $A$ are defined as
$$ \bar{\bm{\rho}} \triangleq \frac{1}{A}\int \bm{x}\, \mathrm{d}\alpha \qquad \text{ and }\qquad A \triangleq \int \, \mathrm{d} \alpha $$In the plane one often defines
$$ \bm{J} \triangleq \int \hat{\bm{r}} \otimes \hat{\bm{r}} \, \mathrm{d}\alpha $$In terms of $\bm{I}$ this is
$$ \bm{J} = - \mathbf{i}^{\times}\bm{I}\mathbf{i}^{\times} $$and
$$ \bm{I} = I_0 \mathbf{i}\otimes \mathbf{i} - \int \hat{\bm{r}}^{\times} (\mathbf{i}\otimes\mathbf{i})\hat{\bm{r}}^{\times} $$or
$$ \bm{I} = I_0 \mathbf{i}\otimes \mathbf{i} - \mathbf{i}^{\times}\int \hat{\bm{r}}\otimes\hat{\bm{r}}\,\mathrm{d}\alpha \, \mathbf{i}^{\times} $$so that
$$ \bm{I} = I_0 \mathbf{i}\otimes \mathbf{i} - \mathbf{i}^{\times} \bm{J} \, \mathbf{i}^{\times} $$It is convenient to note that with coordinates \(\bm{r} = [x, y, z]\)
$$ [\bm{r}^\times][\bm{r}^\times]^{\mathrm{t}} =-\left[\begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{array}\right]^2=\left[\begin{array}{ccc} y^2+z^2 & -x y & -x z \\ -y x & x^2+z^2 & -y z \\ -z x & -z y & x^2+y^2 \end{array}\right] $$