Rectangle
Parameterizing the Surface
We can do this with just standard Cartesian coordinates, so the parameterization is very easy.
\[\b{r}(x, y) = x \; \u{x} + y \; \u{y} \\[10pt] A = \{ \b{r}(x, y) \ : \ -a/2 \leq x \leq a/2 \quad -b/2 \leq y \leq b/2 \}\]Therefore
\[\frac{\partial \b{r}}{\partial x} = \u{x} \qquad \frac{\partial \b{r}}{\partial y} = \u{y}\] \[d \b{A} = \left ( \frac{\partial \b{r}}{\partial x} dx \right ) \times \left ( \frac{\partial \b{r}}{\partial y} dy \right ) = dx \; dy \; \u{z}\] \[dA = \abs{ d \b{A} } = dx \; dy\]Mass
\[\begin{align} M &= \int \; dm \\[10pt] &= \sigma \int \; dA \\[10pt] &= \sigma \int_{-a/2}^{a/2} \int_{-b/2}^{b/2} dx \; dy \\[10pt] &= \sigma \left ( \int_{-a/2}^{a/2} dx \right ) \left ( \int_{-b/2}^{b/2} dy \right ) \\[10pt] &= \sigma (a) (b) \\[10pt] &= \sigma \cdot ab \end{align}\]Moment of Inertia About Central Axis
Moment of Inertia About Central Diameter
Inertia Tensor
\[I = \begin{bmatrix} \tfrac{1}{12} M b^2 & 0 & 0 \\ 0 & \tfrac{1}{12} M a^2 & 0 \\ 0 & 0 & \tfrac{1}{12} M (a^2 + b^2) \end{bmatrix} = \tfrac{1}{12} M \begin{bmatrix} b^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2 + b^2 \end{bmatrix}\]Interestingly, this is exactly the same as the flat elliptical surface, but with the coefficient $\tfrac{1}{12}$ instead of $\tfrac{1}{4}$.