Does anyone know how to determine the moment of inertia of the filled circular sector shown below?
Apparently I have to use $dA=r\ d\theta\ dr$, $y= r\sin(\theta)$ and end up getting $I_x=\frac{r^4}{8}(\alpha-\sin(\alpha))$
But I have no idea how they got that, I have never had this. Could someone step by step explain how to get to $I_x$ with this question and how to integrate something that has two different differentials.
Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.
To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a right triangle with a circular hypotenuse. Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).
The moment of inertia is given by $$ I_1=\rho\iint y^2\ dydx, $$ where $\rho$ is the mass density per unit area, which looks simple enough. The difficulty is just in getting the correct limits of the double integral. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. This gives $$ \begin{align} I_1&=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\ &=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\ &=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2) \end{align} $$ This simplifies to $\frac{\rho}{12} bh^3$ which is the well known value for the moment of inertia of a triangle.
This one goes the same way as the last one, but the limits and the integration are more difficult. This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$. Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$. Putting this into the integral gives $$ \begin{align} I_2&=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\ &=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\ &=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right) \end{align} $$ That last integral was quite tricky, and I ended up just plugging it into Mathematica. I'm sure it is possible to find it in the standard integral tables though.
Finally, putting everything together and working through some trig identities simplifies the whole thing to $$ 2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha) $$ Which is again what we expect from the standard table for a circular segment.
Consider an infinitesimal element of area $r d\theta dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis.
Its moment of inertia is $r d\theta dr (r \sin (\theta ))^2$.
The moment of inertia about the $x$ axis of the complete sector:
$$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{1}{8} r_0^4 (\alpha -\sin (\alpha ))$$
