Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV resit / 3

At a regular point of the parametrized surface, the partial derivatives ${\mathbf{r}}_s$ and ${\mathbf{r}}_t$ are linearly independent tangent vectors. Their cross product gives a nonzero normal vector:Dividing by its Euclidean norm gives a unit normal; changing the sign chooses the opposite orientation. If the cross product vanishes, this parametrization is singular there and cannot determine a normal vector by this formula.

An infinitesimal parameter rectangle becomes a parallelogram with edges ${\mathbf{r}}_{s}ds$ and ${\mathbf{r}}_{t}dt$. Its area is the Euclidean norm of their cross product, giving the surface area elementThus, for a regular parametrization covering the surface once except possibly on sets of area zero,For several patches, add their surface integrals while avoiding overlap; repeated coverage would otherwise count area with multiplicity.

The partial derivatives of this parametrized surface areTheir cross product isThe mixed terms cancel in its squared Euclidean norm:Since $s>0$, the surface area element is $s(t^2+2)dsdt$. There is no repeated coverage: $z=2s$ recovers $s$, and $\sqrt{x^2+y^2}=st$ then recovers $t$. Therefore the surface area is

Use the polar radius $r=at\in [0,1]$ as the parameter of the Archimedean spiral. Its angle in polar coordinates is $r/a$. The radial and angular contributions to its arc length are orthogonal, soThere is no need to evaluate this integral. The simple bounds $r\le \sqrt{a^2+r^2}\le r+a$ implyThe squeeze theorem therefore gives the scaled length of an Archimedean spiral:Geometrically, the radial contribution becomes negligible after multiplying the arc length by $a$; the remaining contribution is the area under the function graph $r\mapsto r$ on $[0,1]$.