Moment of Inertia of a sector of a circle I am trying to find the moment of inertia about its centre of a sector of a circle of radius $a$, mass $m$ and angle $\pi/3$. I have found the answer it is $\frac{1}{2}ma^2$ but originally tried a method that was unsuccessful and don't know why this is the case. This involved finding the moment of inertia of elementary rods about the centre, length $a$ and width $a\;\delta\theta$, using $p$ as mass per unit area. I then tried to integrate the moment of inertia of these rods from $\theta=0$, to $\theta=\pi/3$. Which gave me an answer that did not lead to the correct result. Please can you explain why? I have shown my workings below but they are not very clear: 

 A: When integrating in (2D) polar coordinates you need to use a surface element:
$$dA = r\;dr\;d\theta$$
The reason for this can be seen geometrically:

The surface element has the same shape as one of the spaces between two red and two blue lines (a sort of curved rectangle). In the infinitesimal limit the area of one such segment is just its length multiplied by its width. The length is easy, it is $dr$, and is always the same (notice that the length of a blue segment between the evenly spaced red lines is always the same). The width is a bit more subtle. First, you might notice that the "inner width" and "outer width" are different. We don't need to worry about this because in the infinitesimal limit, they approach the same length. But the length of the arc between two evenly spaced blue lines clearly increases as the radial coordinate increases. It should be easy to convince yourself that the length of an arc that spans angle $theta$ is $r\theta$ (of course $theta$ in radians). It follows that the width of the surface element is $r\;d\theta$ (the angle shrinks to an infinitesimal, but the radial coordinate does not - it simply takes the value of the radius wherever we place our surface element).
So why doesn't your integral using a line work? Well, this treats surface elements at all distances from the origin as having the same size (I'm speaking very loosely here, since infinitesimals don't really have a size). But looking at the diagram, clearly surface elements close to the origin need to be smaller than ones further away, otherwise you end up "overcounting" area near $r=0$ and under-counting area further out.
Put another way, take a bunch of long thin rectangles of the same size cut out of paper and try to arrange them into an approximate segment of a circle without overlapping them. You should find it's impossible. Even as you make them infinitesimally thin this fails. You need to use little wedge shapes pieces.
