Let us consider a very small volume element PQRSS’P’Q’R as shown in Fig.1. The lower surface PQRS of this volume element lies on the surface of the sphere of radius \( r \) and the parallel surface P’Q’R’S’ lies on the surface of the sphere of radius \( (r+dr) \). Since the angles \( d\theta \) and \( d\phi \) and radial extension \( dr \) are very small, so the volume element can be assumed as rectangular volume element.

Now the vectors \( \vec{PS} \), \( \vec{PQ} \) and \( \vec{PP’} \) represent the three side of the elemetary volume lement PQRSS’P’Q’R‘. From Fig. 1 we can write,

\( \vec{PS}=r\ d\theta\ \hat{\theta} \)

\( \vec{PQ}=r\ \sin\theta\ d\phi\ \hat{\phi} \) and

\( \vec{PP’}=dr\ \hat{r} \)

So the volume of the elementary volume element PQRSS’P’Q’R’ is given by,

\( dV=\vec{PP’}\cdot(\vec{PS}\times\vec{PQ}) \)

\( or,\ dV=[dr\ \hat{r}]\cdot[(r\ d\theta\ \hat{\theta})\times(r\ \sin\theta\ d\phi\ \hat{\phi})] \)

\( or,\ dV=[dr\ \hat{r}]\cdot[r^2\ \sin\theta\ d\theta\ d\phi\ \hat{r}] \)

\( or,\ dV=r^2\ \sin\theta\ dr\ d\theta\ d\phi[\hat{r}\cdot\hat{r}] \)

\( or,\ dV=r^2\ \sin\theta\ dr\ d\theta\ d\phi \)

where, unit vectors \( \hat{r} \), \( \hat{\theta} \) and \( \hat{\phi} \) are constituted in a right handed orthogonal co-ordinate system, as a result, \( \hat{\theta}\times\hat{\phi}=\hat{r} \) and \( \hat{r}\cdot\hat{r}=1 \).

This volume is a scalar quantity.