Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / Section B / 6

Taking the curl component by component gives

Integrating ${\psi }_x=-(2xy+z)$ with respect to $x$ gives $\psi =-x^{2}y-xz+h(y,z)$. Matching ${\psi }_y=-(x^2+2y)$ requires $h=-y^2+j(z)$, and matching ${\psi }_z=-x$ requires $j$ constant. A scalar potential with the requested sign isThus the force field is a conservative vector field on all of ${\mathbb{R}}^3$.

Parametrize the straight line by $\mathbf{x}(t)=(t,t,t)$, $0\le t\le 1$. The line integral isThe change in scalar potential is $\psi (1,1,1)-\psi (0,0,0)=-3$, so the answer is $-\Delta \psi$, as required by the fundamental theorem for line integrals for $\mathbf{F}=-\nabla \psi$.

For positive semiaxes $a,b$, the cross product of the two parametric tangent vectors gives the outward oriented surface element of an ellipsoid of revolution:Substituting the parametrization into the vector field and taking its dot product with this element givesThe middle term has zero integral over $0\le \varphi \le 2\pi$. In the other terms, ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$, ${\int }_0^{\pi }{\sin }^3\theta {\cos }^2\theta d\theta =4/15$, and ${\int }_0^{\pi }\sin \theta {\cos }^4\theta d\theta =2/5$. The directly evaluated flux is therefore