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Differentiate the position vector with respect to each of the spherical polar coordinates. The derivatives areTheir lengths give the scale factors of orthogonal coordinates:Dividing the derivatives by these lengths yieldsEach has unit length. For example, ${\mathbf{e}}_r\cdot {\mathbf{e}}_{\theta }=\sin \theta \cos \theta ({\cos }^2\varphi +{\sin }^2\varphi )-\cos \theta \sin \theta =0$; 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 $r=0$ and on the polar axis, but this does not affect an area integral.

On the parametrized surface $\theta =\alpha$, use $r,\varphi$ as parameters. The surface integral area element isFor $0\le \alpha \le \pi$, its lateral area, with no closing disk, is thereforeAt $\alpha =0,\pi$ the surface collapses to a line and its area is zero.

Lagrange's theorem says that for a finite group $G$ and a subgroup $H$,In particular, $|H|$ divides $|G|$. To prove it, consider the left cosets $gH$. Every element lies in one, and if $gH$ and $g^{\prime }H$ intersect, then $gh=g^{\prime }{h}^{\prime }$ for some $h,h^{\prime }\in H$. Hence $g=g^{\prime }{h}^{\prime }{h}^{-1}$ and $gH=g^{\prime }H$. Distinct cosets are therefore disjoint and partition $G$. Multiplication by $g$ is a bijection $H\to gH$, so every coset has $|H|$ elements. Counting this partition proves the formula.

Write the dihedral group aswhere $r$ is a rotation and $s$ a reflection. If $k\mid n$, the cyclic subgroup $⟨r^{n/k}⟩$ has order $k$. If $k\mid 2n$ but $k\nmid n$, then $k$ is even: an odd divisor of $2n$ would divide $n$. Put $m=k/2$, so $m\mid n$. The subgroupcontains exactly the $m$ rotations $r^{jn/m}$ and the $m$ reflections $r^{jn/m}s$, for $0\le j<m$. The inversion relation makes this set closed under multiplication and inverses. These $2m=k$ elements are distinct, so it has the required order. This proves the dihedral subgroups of every divisor order property. The construction works for every even divisor, including those already covered by the cyclic construction.

Use suffix notation, with repeated indices summed. Contracting the two Levi-Civita symbols givesApplying the product rule to $h=A_i{B}_i$ and substituting the two evolution equations yieldsIn the second line we renamed dummy indices in $A_i{B}_j{\partial }_j{u}_i$ and then combined product derivatives. HenceNo zero-divergence assumption was used in this local identity.

When both fields are solenoidal vector fields, the product rule for divergence rewrites this as a conservation law:Because the volume is fixed, differentiation under its integral and the divergence theorem giveunder the two tangency conditions. Thus the volume integral is conserved. This is helicity conservation by tangent boundary conditions; when $\mathbf{B}=\nabla \times \mathbf{A}$, the density is the magnetic helicity density.

For the given axisymmetric field, the curl in spherical coordinates givesThere are no $\varphi$ derivatives and $A_r=0$. ConsequentlyThe volume form in spherical polar coordinates is $r^2\sin \theta drd\theta d\varphi$, soHere $B_r=0$ at $r=a$, as required for the magnetic-field tangency condition in the conservation argument; tangency of $\mathbf{u}$ is a separate hypothesis.

The divergence theorem states that a continuously differentiable vector field on a bounded region with piecewise smooth boundary satisfieswith the unit normal pointing out of $V$. Apply it separately to $\mathbf{u}=f{\mathbf{e}}_i$, where ${\mathbf{e}}_i$ is each constant Cartesian basis vector. Since $\nabla \cdot (f{\mathbf{e}}_i)={\partial }_{i}f$ and $(f{\mathbf{e}}_i)\cdot \mathbf{n}=f{n}_i$, the three component identities combine to give

For the specified scalar field, $\nabla f=(z,0,x)$. The $x$ term integrates to zero by reflection symmetry. At height $z$, the region has a disk cross-section of area $\pi (a^2-cz)$, and $0\le z\le a^2/c$. Therefore the volume integral isTo check the surface integral independently, split the boundary into the base disk and the curved graph $z=g(x,y)=(a^2-x^2-y^2)/c$. On the base $f=0$, so its contribution vanishes. The region lies below the curved graph, whose outward vector surface element of a graph isOver the disk $x^2+y^2\le a^2$, the $y$ and $z$ components of $fd\mathbf{S}$ are odd in $x$ and integrate to zero. Its $x$ component gives, in polar coordinates,Thus the two explicitly evaluated integrals agree, with the outward orientation included.