Question:
Determine the moment of inertia for a quadrant of a uniform circular lamina of radius b.
Here I saw the answer that,however I don't understand it first of all here is the answer and I will post what I don't understand about it.
$I_z = \int R^2 \rho ds = \int_0^b \int_{-\pi/4}^{\pi/4} sin^2(\theta) r^3 \rho d\theta dr$
Alright first of all why did we take the moment of inertia about the z-axis why not y or x. Secondly I don't understand the bounds or specifically for calculation of $d\theta$ can someone clarify ??
If body is rotating around specific axes, quantity you are looking for is called angular momentum of inertia.
$L = r \times mv = I\omega e_z = I_z\omega$, where $r$ - is a vector from rotation axe to the point, $v$ is a vector of speed, $m$ is a inertial ass of the point, $\omega$ - angular velocity and $\omega r = v$,$m = \rho s$ where $\rho$ - is an area density and $s$ is area of "point" of mass. Because rotation is happening in x-y plane, vector product is going to be along z-axe.
Switching to polar coordinates for the circle wedge we are having: $$I_z = \int r^2 dm = \int r^2 \rho ds = \int_0^b \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}r^3\rho drd\theta$$ So there is no way $\sin(\theta)$ comes from ... unless some additional conditions are missing.
UPDATE
The only way you can get $\sin^2(\theta)$ under integral if disk has it thinkness or density proportional to $\sin^2(\theta)$ ... so it will be not a disk but some strange object.