In spherical polar co-ordinate system, there are three surface elements \( {dA}_r,\ {dA}_{\theta} \) and \( {dA}_{\phi} \).

[To know about the three types of area elements and their values (CLICK HERE)]

Now in case of the surface element \( {dA}_r \),  \( r \) is constant but \( \theta \) and \( \phi \) are variable.

\( {d\vec{A}}_r=r^2\ \sin\theta\ d\theta\ d\phi\ \hat{r} \)

And the magnitude is \( {dA}_r= r^2\ \sin\theta\ d\theta\ d\phi \)

Here, the area element is perpendicular to the radial vector \( \vec{r} \), i.e., the direction of the area element is in the same direction of the radial vector \( \vec{r} \), this means that the angle between \( {d\vec{A}}_r \) and \( \vec{r} \) is zero.

So the solid angle subtended by this area element at the origin is given by,

\( d\omega=\frac{dA\ \cos{0}^\circ}{r^2}=\frac{ r^2\ \sin\theta\ d\theta\ d\phi }{r^2} \)

\( or,\ d\omega=\sin\theta\ d\theta\ d\phi \)

Now in case of the surface element \( {dA}_{\theta} \), \( \theta \) is constant but \( r \) and \( \phi \) are variable. So the area element is given by,

\( {d\vec{A}}_{\theta}=r\ \sin\theta\ d\phi\ dr\ \hat{\theta} \)

So the direction of this area element \( {d\vec{A}}_{\theta} \) is along \( \hat{\theta} \), which is perpendicular to the direction of the radial vector \( \vec{r} \).

Therefore the component of the area element \( {d\vec{A}}_{\theta} \) along the direction of the increasing \( \vec{r} \) is given by,

\( |{d\vec{A}}_\theta|\ \cos{90}^\circ=0 \),

This indicates that the area element \( {d\vec{A}}_\theta \) is co-planar with \( r \).

So the solid angle subtended by this area element at the origin is 0.

Now in case of the surface element \( {dA}_{\phi} \), \( \phi \) is constant but \( r \) and \( \theta \) are variable. So the area element is given by,

\( {d\vec{A}}_{\phi}=r\ dr\ d\theta\ \hat{\phi} \)

So the direction of this area element \( {d\vec{A}}_{\phi} \) is along \( \hat{\phi} \), which is perpendicular to the direction of the radial vector \( \vec{r} \).

Therefore the component of the area element \( {d\vec{A}}_{\phi} \) along the direction of the increasing \( \vec{r} \) is given by,

\( |{d\vec{A}}_\phi|\ \cos{90}^\circ=0 \).

This indicates that the area element \( {d\vec{A}}_\phi \) is co-planar with r.

So the solid angle subtended by this area element at the origin is 0.

Now It is clear that, in the spherical polar co-ordinate, one type of area element \( {d\vec{A}}_r \) subtends a solid angle at the origin.