If we have curve $r=F(\varphi)$ and two polar radii $\varphi=\alpha$ and $\varphi=\beta$ $(\alpha<\beta)$. Then volume of body which is bounded by above polar curve and lines which is revolved around polar axis is equal to the following integral: $$\dfrac{2\pi}{3}\int \limits_{\alpha}^{\beta}r^3\sin \varphi \ d\varphi$$
This is from my calculus Russian book. How to derive that formula?
Can anyone please show it?
The volume of a body in spherical coordinates is given by
$$V=\int_V dV=\int_{\varphi_1}^{\varphi_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}r^2\sin\varphi~dr~d\theta~d\varphi$$
Now, if $r=r(\varphi)$ only and it is a body of revolution, i.e. $\theta\in[0,2\pi]$, then
$$ \begin{align} V &=2\pi~\int_{\varphi_1}^{\varphi_2}\int_{r_1}^{r_2}r^2\sin\varphi~dr~d\varphi\\ &=2\pi~\int_{\varphi_1}^{\varphi_2}\frac{r^3}{3}\sin\varphi~d\varphi\\ &=\frac{2\pi}{3}~\int_{\varphi_1}^{\varphi_2}r^3\sin\varphi~d\varphi\\ \end{align} $$
as expected.