= Solution
Write $q=\mathbf u\cdot\boldsymbol\omega$ for the <kinetic helicity density>, and retain $\mathcal B=w+u^2/2$. Since $\boldsymbol\omega$ is a <curl>, $\nabla\cdot\boldsymbol\omega=0$. Differentiating the density and using the equations from the preceding part gives
$$
\partial_tq=-\boldsymbol\omega\cdot\nabla\mathcal B+\mathbf u\cdot\nabla\times(\mathbf u\times\boldsymbol\omega).
$$
The <divergence of a cross product> gives
$$
\nabla\cdot\{\mathbf u\times(\mathbf u\times\boldsymbol\omega)\}=(\mathbf u\times\boldsymbol\omega)\cdot\boldsymbol\omega-\mathbf u\cdot\nabla\times(\mathbf u\times\boldsymbol\omega).
$$
The first term on the right is zero, while $\boldsymbol\omega\cdot\nabla\mathcal B=\nabla\cdot(\mathcal B\boldsymbol\omega)$. Thus the <kinetic helicity conservation law> has flux
$$
\boxed{\mathbf F_{H_k}=\mathcal B\boldsymbol\omega+\mathbf u\times(\mathbf u\times\boldsymbol\omega)=q\mathbf u+\left(w-\frac{u^2}{2}\right)\boldsymbol\omega,\qquad\partial_tq+\nabla\cdot\mathbf F_{H_k}=0.}
$$
The last equality for the flux uses the <vector triple product identity>. Integrating and applying the <divergence theorem> gives $d\int_Vq\,dV/dt=-\int_{\partial V}\mathbf F_{H_k}\cdot\mathbf n\,dS$ for a fixed volume. Hence its total <kinetic helicity> is conserved when the boundary flux vanishes; for example, tangency of both <velocity field> and <vorticity> to the boundary is sufficient. The local <conservation law> does not by itself imply that its density follows each fluid particle unchanged.
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