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Prove that the volume of the sphere is $\frac{4}{3}\pi R^3$ We can think of a sphere as made up of infinite number of concentric shells of thickness $dr$ and volume $4\pi r^2dr$ So the total volume of sphere is equal to $$\int_0^R4\pi r^2dr = 4\pi[\frac{r^3}{3}]_0^R = \frac{4}{3}\pi R^3$$ Is this correct ?

Andrei
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user677491
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  • basically correct, but I think it would be better to use two different letters (e.g., $R$ and $r$) instead of using $r$ for both the radius of the sphere and the variable of integration – J. W. Tanner Jul 05 '19 at 13:56
  • Yes, modulo permission to use the differential $dr$ to approximate the volume of the shell assuming you know its surface area. – Ethan Bolker Jul 05 '19 at 13:56
  • as per your derivation the sphere you are considering is a solid non conducting sphere (as it is said in most elementary electrostatics textbooks) But for spherical shell your method won't work. There you have to take either zonal strip or work via spherical co-ordinate system to get the result.Since interior is hollow you won't get concentric shells in 'air' which you got inside solid sphere –  Jul 05 '19 at 14:21

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Yes, it is correct. You can verify it intuitively, as an expanding spherical shell whose world space is the solid sphere.