This is the exercise im struggling with:
The surface density of a thin rectangle varies as: $\sigma(x,y)= 11.0$ kg/m$^2$ $+ 5.00 $ kg/m$^4 (x^2+y^2)$ The rectangle has a length $L = 0.900$ m and a width $W = 0.900$ m. Calculate $I_z$, the moment of inertia about the $z$-axis.
Use the integral for the moment of inerzia around the $z$ axis:
$$I_z = \int \ell^2 dm$$
Where $dm$ is the mass element and $\ell$ is the distance from the axis.
In your case you have a rectangle (which is actually a square, since width and length are equal). The distance is then half the diagonal:
$$\ell^2 = d^2 = \left(\frac{x}{2}\right)^2 + \left(\frac{y}{2}\right)^2 = \frac{x^2 + y^2}{4}$$
$$dm = \sigma dS$$
Where $\sigma$ is your surface density and $dS = dx dy$ is the area element. Hence the integral becomes
$$\int_0^{0.9} dx \int_0^{0.9} dy \frac{x^2 + y^2}{4} (11.0 + 5(x^2 + y^2)) dx dy$$
The integral is pretty trivial, do the maths and you will get
$$\frac{1}{4}\left(\frac{11}{3}x^3 + \frac{11}{3}y^3 + x^5 + y^5 + \frac{2}{9}x^3y^3\right)$$
To be calculated at $x = y = 0.9$.