# 3D Dynamics: determining the moments of inertia on a plate with a couple

My Mechanical textbook (Bedford & Fowler 4th Edition) has a worked out example for determining a couple using euler's equations. This is not a homework question (at least I don't think it is?), this is a request for some elaboration on a pre-worked example.

I am trying to work through the problem myself to come up with the same answer but I have inconsistencies with the values for the moment of inertia tensor. Here are the relevant pages of the book (apologies for the lack of quality on the pictures):

As shown on page 2, the moment of inertia tensor is calculated as: $$\begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{yx} & I_{yy} & -I_{yz} \\ -Izx & -Izy & Izz \\ \end{bmatrix} = \begin{bmatrix} 0.48 & -0.18 & 0 \\ -0.18 & 0.12 & 0 \\ 0 & 0 & 0.6 \\ \end{bmatrix}$$

The values for the moments of inertia about the x, y and z axes for the plate seems to conflict with my understanding:

$$I_{xy} = \int_{}^{}f(xy)\,dm = (0.150 * 0.300) * 4 kg = 0.18 kg.m^2$$ $$I_{yx} = \int_{}^{}f(yx)\,dm = (0.300 * 0.150) * 4 kg = 0.18 kg.m^2$$ $$I_{xz} = \int_{}^{}f(xz)\,dm = (0.150 * 0) * 4 kg = 0.00 kg.m^2$$ $$I_{zx} = \int_{}^{}f(zx)\,dm = (0 * 0.150) * 4 kg = 0.00 kg.m^2$$ $$I_{zy} = \int_{}^{}f(zy)\,dm = (0 * 0.300) * 4 kg = 0.00 kg.m^2$$ $$I_{yz} = \int_{}^{}f(yz)\,dm = (0.300 * 0) * 4 kg = 0.00 kg.m^2$$ $$I_{zz} = \int_{}^{}f(x^2 + y^2)\,dm = (0.150^2 + 0.300^2) * 4 kg = 0.45 kg.m^2$$ $$I_{xx} = \int_{}^{}f(y^2 + z^2)\,dm = (0.300^2 + 0^2) * 4 kg = 0.36 kg.m^2$$ $$I_{yy} = \int_{}^{}f(x^2 + z^2)\,dm = (0.150^2 + 0^2) * 4 kg = 0.09 kg.m^2$$

Thus my inertia tensor is different from the book's: $$\begin{bmatrix} 0.36 & -0.18 & 0 \\ -0.18 & 0.09 & 0 \\ 0 & 0 & 0.45 \\ \end{bmatrix} \ne \begin{bmatrix} 0.48 & -0.18 & 0 \\ -0.18 & 0.12 & 0 \\ 0 & 0 & 0.6 \\ \end{bmatrix}$$ Could someone explain to me what conceptual mistake I have made?

For a uniform plate with the reference rotation being at any corner, shouldn't the x,y and z distance (for the moment of inertia of the respective axis) be taken from the center of gravity of the plate to the rotation point?

Be careful with your integration elements, as $dm=\rho dA$, where $\rho$ is the area density of the plate and $dA=dxdy$ is the area integration element. In addition you should integrate from the rotation point to the end of your plate. So for example;
$$$$I_{xx}\\=\int_A(y^2+z^2)\rho dA\\=\rho\int_{x=0}^{0.3\text{ m}}dx\int_{y=0}^{0.6\text{ m}}y^2dy\\=\frac{4\text{ kg}}{0.3\text{ m}\cdot0.6\text{ m}}\cdot x|_0^{0.3\text{ m}}\cdot\tfrac{1}{3}y^3|_0^{0.6\text{ m}}=0.48 \text{ kg}\,\text{m}^2.$$$$
Note that maybe by accident the error you made in the $I_{xy}$ term cancels
$$I_{xy}=I_{yx}\\=\int_Axy\rho dA\\=\rho\int_xxdx\int_yydy\\=\rho\tfrac{1}{2}x^2|_0^{0.3}\tfrac{1}{2}y^2|_0^{0.6}=0.18\text{ kg}\,\text{m}^2.$$