How does $dV=2\pi R^2t \sin \theta\, d\theta$ come?

Question

How does $dV=2\pi R^2 t \sin \theta\, d\theta$ come?

From the figure shown in the screenshot, How do I prove $dV=2\pi R^2 t \sin \theta \,d\theta$?

My Attempt

Answer 1

These are differentials. We are allowed to multiply the elemental volumes just as for rectangular / cuboid blocks. Cone differential area with $ds$ along local cone tangent

$$dA= 2\pi r ds = (2\pi R \sin \theta) R d\theta$$

For constant shell thickness

$$ dV = t\cdot dA$$

The sphere has this special property..

The truncated spherical segment volume equals segment height $h$ times $2 \pi R$ i.e., should be multiplied by $ R(\cos \theta_2-\cos \theta_2)=h,$

For differential volume next multiply by shell thickness $t$.

First question: Area by definition of a curvilinear surface is the rate at which the Volume is increasing... If you differentiate $\frac43 \pi R^3$ with respect to $R$ you get the area $ 4 \pi R^2.$

Second question $ dV_{inside\;of\;ball}= \pi r^2 dz$ is true for the volume inside a soccer ball for whereas $dV_{thick\;shell\;of\;ball\;}=t\cdot dA. $

Answer 2

You can calculate it with a Jacobian determinant or differential forms, but the diagram allows a simpler argument. The shaded ring has circumference $2\pi R\sin\theta$, which we multiply by the width and thickness to get the stated result.