Differentiate the position vector with respect to each of the spherical polar coordinates. The derivatives are
Their lengths give the scale factors of orthogonal coordinates:
Dividing the derivatives by these lengths yields
Each has unit length. For example, ; the other two dot products vanish because their first two terms cancel. Thus these are mutually orthogonal vectors, and the total differential has the required form. The coordinates degenerate at and on the polar axis, but this does not affect an area integral.
On the parametrized surface , use as parameters. The surface integral area element is
For , its lateral area, with no closing disk, is therefore
At the surface collapses to a line and its area is zero.

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