Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 11 / a

The gradient part contributes zero around a closed path by the fundamental theorem for line integrals:Along the parametrization, $x^2+z^2=1$, so the first two components of $\mathbf{w}$ vanish wherever the printed expression is defined. The remaining component gives $\mathbf{w}\cdot d\mathbf{x}=xdz={\cos }^2\theta d\theta$. ThusThe printed field needs a qualification: $\mathrm{arctanh}(x^2)$ is undefined at the isolated parameters where $x^2=1$, so $0\mathrm{arctanh}(1)$ cannot literally be evaluated. The result above is the improper line integral, or equivalently the integral of the continuous extension of the restricted field along the curve. That restricted field tends to $(0,0,\cos \theta )$ at each omitted parameter. No extension of the full field to a neighbourhood is being assumed.

The curl of a gradient vanishes because the mixed second partial derivatives of the smooth function $\varphi$ agree. Therefore $\nabla \times \mathbf{v}=\nabla \times \nabla \varphi =0$ everywhere, andUnlike part (i), this integrand does not involve the problematic auxiliary field.

Where $\mathbf{A}=|\mathbf{v}{|}^2\mathbf{w}$ is twice continuously differentiable, the identity divergence of a curl is zero gives $\nabla \cdot (\nabla \times \mathbf{A})=0$. All regular portions of the parameter interval lie in this region, henceas an improper line integral. Besides the $\mathrm{arctanh}(x^2)$ endpoint problem from (i), the curve touches $y=1$ at $\theta =\pi /2$, where the printed real $\arcsin y$ expression has no smooth open neighbourhood. The identity applies on the intervening open segments, and its zero value has a unique continuous extension along the curve. A literal classical integrand at every parameter is not supplied by the printed field.