Volume of a sector of a solid sphere
I am trying to solve a question in Dynamics which involves finding volume of a sector of a solid sphere.
Here, the $\mathrm{d}V$ they are using for finding the volume is $\mathrm{d}V = 2\pi R \sin(\theta) R\ \mathrm{d}\theta \mathrm{d}R$.
Can anyone please explain what element they are using. I think they are using some sort of ring. My next question is if they are using a ring, how is it able to give us correct answer since the sphere is not hollow!
The volume element is $dV=R^2\sin(\theta)dRd\theta d\varphi$, in spherical coordinates. The spherical coordinates can be taken to be $$x=R\sin(\theta)\cos(\varphi)$$ $$y=R\sin(\theta)\sin(\varphi)$$ $$z=R\cos(\varphi),$$ where $\theta$ is the angle of the vector $(x,y,z)$ with the positive $z$-axis (taking values from $0$ to $\pi$) and $\varphi$ is the angle the vector $(x,y,0)$ makes with the positive $x$ axis measured counterclockwise (taking values from $0$ to $2\pi$). You can check that the Jacobian gives the value $R^2 \sin(\theta)$.
In this case, since we are removing a spherical sector, the sector will be of the form $$\sigma:=\{(R,\theta,\varphi):0\leqslant R\leqslant a,0\leqslant \theta\leqslant \alpha,0\leqslant \varphi\leqslant 2\pi\}.$$ That is, when we parametrerize the sector removed, because it is rotationally symmetric about the $z$-axis, it will involve all values of $\varphi$ in $[0,2\pi]$. We also have an integrand (call it $f$) which does not depend on $\varphi$. So we should integrate $$\int_0^{2\pi}\int_0^a\int_0^\alpha f(R,\theta) R^2\sin(\theta) d\theta dRd\varphi.$$ Since the integrand does not depend on $\varphi$, the $\int_0^{2\pi}\ldots d\varphi$ is what yields the $2\pi$ and causes the $d\varphi$ to disappear.