Rotational Angular Momentum of a Rigid Body - a Derivation Difficulty I was attempting a derivation of the angular momentum of a rigid body (rotational only, no translational) using purely a vector-based approach.
In the image attached, I have set the centre of mass of the rigid body in 3 dimensions at the origin of the cartesian coordinate system. The resolution of each particle's position vector into a component perpendicular and parallel to the axis of rotation (z axis) was used to the effect of achieving nice cancellation of other terms.
At the end, I am left with the expected I(omega) as well as another term which seems impossible to see vanish. Why does this additional term appear here? I'd appreciate some direction as to how and why I'm getting this erroneous result.
Thanks in advance! Any help is greatly appreciated!

 A: Congratulations!  You have just discovered the inertia tensor.  As a general rule, the angular velocity is related to the angular momentum not simply by $\vec{L} = I \vec{\omega}$, but by the matrix equation
$$
\begin{bmatrix} L_x \\ L_y \\ L_z \end{bmatrix} = \underbrace{\begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{bmatrix}}_{\equiv \mathbf{I}}\begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \end{bmatrix},
$$
where the on-diagonal components of the matrix $\mathbf{I}$ are called the moments of inertia and are given by
$$
I_{zz} = \sum_i m_i (x_i^2 + y_i^2)
$$
(and similarly for $I_{xx}$ and $I_{yy}$), and the off-diagonal components are called the products of inertia are given by
$$
I_{xy} = -\sum_i m_i x_i y_i
$$
(and similarly for $I_{yx}$, $I_{xz}$, etc.) Note that this latter definition implies that this matrix is symmetric.
In your case, where $\omega = \omega_z \hat{z}$, the term you wanted is given by $L_z = I_{zz} \omega_z$, while the extra unexpected terms are $L_x = I_{xz} \omega_z$ and $L_y = I_{yz} \omega_z$.  So those "spurious" terms are really supposed to be there, and your result is entirely concordant with this picture.
In general, the relationship between a body's angular momentum and its angular velocity is given by this matrix equation.  However, there is a special choice of axes, called the principal axes, in which all of the off-diagonal components of $\mathbf{I}$ are zero.  In this case, we have $L_x = I_{xx} \omega_x$, and similarly for the $y$- and $z$-axes.  These principal axes can be found by the process of diagonalizing the inertia tensor; in general, though, they often coincide with any axes or planes of symmetry the body has.  In particular, most of the "nice" axes for which you learned the moments of inertia in first-year physics were pre-selected to be principal axes of the body.  This makes the lives of first-year physics students easier, but does lead to a surprise when they learn about the full story later on.
If you want more information about this, it is covered in most intermediate-level mechanics textbooks.  I can recommend Taylor's Classical Mechanics, Thornton & Marion's Classical Dynamics of Particles & Systems, or Kleppner & Kolenkow's An Introduction to Mechanics.. You can also Google any of the phrases that I put in italics in my description above.
