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Suppose that we have a thin plate, so thin that it's practically 2dimensional. Such a plate is called a planar lamina. A lamina is described by the region $D$ in the $x$$y$ plane that it covers, and by its mass density $\rho(x,y)$, which gives the mass per unit area. In the following video, we show how to get the mass and centerofmass of a lamina by integration. If the density were a constant, finding the total mass of the lamina would be easy: we would just multiply the density by the area. When the density isn't constant, we need to integrate instead. The mass of a little box of area $dA$ around the point $(x,y)$ is essentially $\rho(x,y) dA$. For the total mass of the lamina, we add up the boxes and take a limit to get $$M \ = \ \iint_D \rho(x,y) dA.$$ This integral can be done in rectangular coordinates, polar coordinates, or by whatever method you prefer.
The centerofmass of a body is a weighted average of the positions of the particles inside. Since a box of area $dA$ at position $(x,y)$ has a mass that's a fraction $\rho(x,y) dA/M$ of the total, the centerofmass of our lamina is at position $(\bar x, \bar y)$, where \begin{eqnarray*} \bar x & = & \frac{1}{M} \iint_D x \rho(x,y) dA \\ \bar y & = & \frac{1}{M} \iint_D y \rho(x,y) dA \end{eqnarray*}
The moment of inertia of an object indicates how hard it is to rotate. For a point particle, the moment of inertial is $I=mr^2$, where $m$ is the mass of the particle and $r$ is the distance from the particle to the axis of rotation. The moment of intertia of an object with many pieces is the sum of the moments of inertia of its pieces. The following video what the moment of inertia means physically, and how we can calculate it. Let's imagine that we're rotating around the origin, so $r^2=x^2+y^2$. Since the moment of inertial of a little box of size $dA$ at position $(x,y)$ is $(x^2+y^2) \rho(x,y) dA$, the moment of inertia of the entire lamina is $$ I = \iint_D (x^2+y^2) \rho(x,y) dA.$$
