The surface density of a thin rectangle varies as:

σ(x,y)= 14.0 kg/m2 + 9.00 kg/m4(x2+y2)
The rectangle has a length L = 0.500 m and a width W = 1.500 m. Calculate Iz, the moment of inertia about the z-axis.

I tried this but got a major mess.

x^2+y^2 = r^2 luckily

I = integral r^2(14+9r^2)(2 pi r dr)

but the limits are a serious mess and I was unable to figure it out, sorry

Oh, perpendicular axis theorem might help:

http://hyperphysics.phy-astr.gsu.edu/hbase/perpx.html

To calculate the moment of inertia about the z-axis, we need to integrate the surface density over the area of the rectangle.

First, let's write the expression for the moment of inertia about the z-axis, Iz.

Iz = ∫∫(x^2 + y^2) dm

Next, we need to express the mass dm in terms of the surface density σ(x,y) and the area elements dA.

dm = σ(x,y) * dA

In this case, the surface density is given by σ(x,y) = 14.0 kg/m^2 + 9.00 kg/m^4(x^2 + y^2).

The area element is given by dA = dx * dy.

Now, we can substitute dm and dA into the expression for Iz.

Iz = ∫∫(x^2 + y^2) * (σ(x,y) * dx * dy)

Since the rectangle has a length L = 0.500 m and a width W = 1.500 m, we can set the limits of integration as follows: 0 ≤ x ≤ 0.5 and 0 ≤ y ≤ 1.5.

Now, we can compute the moment of inertia about the z-axis by evaluating the double integral.

Iz = ∫∫(x^2 + y^2) * (14.0 kg/m^2 + 9.00 kg/m^4(x^2 + y^2)) * dx * dy

We can solve the integral in stages, first integrating with respect to x and then with respect to y.

∫(x^2 + y^2) * (14.0 kg/m^2 + 9.00 kg/m^4(x^2 + y^2)) * dx from 0 to 0.5

∫∫(x^2 + y^2) * (14.0 kg/m^2 + 9.00 kg/m^4(x^2 + y^2)) * dy from 0 to 1.5

By evaluating these double integrals, we can find the moment of inertia Iz about the z-axis for the given rectangle.