inertia of a lamina I am given a lamina and the area that it covers is rectangular. Its density, $d$, varies with position: $d=x^2+y^2$. If I want to find the moment of inertia about the $x$ and $y$ axis, can I do the integrals in polar form? I say this because I am aware that I need to (double) integrate $y^2(x^2+y^2)$ to find inertia about $x$-axis, and this isn't easy to do in Cartesian! My lamina is rectangular so polar surely wouldn't work?
Thanks.

It's easier to use Cartesian coordinates,
\begin{align*}
  I_{xx} &= \int_{a}^{b} \int_{c}^{d} \rho y^2 \, dy \, dx \\
  &= \int_{a}^{b} \int_{c}^{d} y^2(x^2+y^2) \, dy \, dx \\
  &= \int_{a}^{b}
     \left[
       \frac{x^2(d^3-c^3)}{3}+\frac{d^5-c^5}{5}
     \right] \, dx \\
  &= \frac{(b^3-a^3)(d^3-c^3)}{9}+\frac{(b-a)(d^5-c^5)}{5} \\
  I_{yy} &= \int_{a}^{b} \int_{c}^{d} \rho x^2 \, dy \, dx \\
  &= \frac{(b^3-a^3)(d^3-c^3)}{9}+\frac{(b^5-a^5)(d-c)}{5} \\
\end{align*}
