Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 13

The field $\mathbf{F}$ is not a conservative vector field: either nonzero closed line integral computed below rules out a scalar potential. Equivalently, its planar curl is ${\partial }_x{F}_y-{\partial }_y{F}_x=1-(-1)=2$.

For $\mathbf{G}$, integrating ${\Phi }_x=2x{y}^2$ gives $\Phi =x^2{y}^2+h(y)$. Matching ${\Phi }_y=2y{x}^2$ then gives $h^{\prime }=0$. ThusThis also explains why both closed line integrals of $\mathbf{G}$ vanish.

Starting at $(1,0)$, traverse the right, top, left and bottom sides in that order. For $\mathbf{F}$, the differential expression is $-ydx+xdy$. The four line integral contributions are respectively $1,1,0,0$, so the total is two.

For $\mathbf{G}$, the expression is $2x{y}^{2}dx+2y{x}^{2}dy$. The right side contributes ${\int }_0^{1}2ydy=1$, and the top side contributes ${\int }_1^{0}2xdx=-1$; the other sides contribute zero. Therefore

Parametrize the circle by $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Its tangent is ${\mathbf{r}}^{\prime }=(-\sin t,\cos t)=\mathbf{F}(\mathbf{r}(t))$, so the dot product $\mathbf{F}\cdot {\mathbf{r}}^{\prime }$ is one. For the other field,Integrating around the full circle gives

Substituting the endpoint parameter values givesThe curve lies on the unit circular cylinder $x^2+y^2=1$. Its azimuth increases from zero to $\pi /2$ while its height increases linearly, so it is one quarter-turn of a rising helix. The arrow in the sketch indicates increasing parameter.

The arc length factor is the norm of the tangent:Hence $ds=10dt$ and $\Phi =xy=\cos 6t\sin 6t=\frac{1}{2}\sin 12t$. The scalar line integral is thereforeThis uses scalar arc length, rather than the oriented displacement used in the vector line integrals of part (a).