Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 3

Integrating the first component with respect to $x$ suggests $\varphi =xy+g(y,z)$. Matching the other components gives $g=\sin (yz)+C$. Direct differentiation verifiesThus the vector field is conservative, with the displayed scalar potential.

The $z$ component of the curl is ${\partial }_x{F}_y-{\partial }_y{F}_x=e^{x+y}-x{e}^{xy}$, which is not identically zero; at the origin it is one. Hence the field is not conservative on ${\mathbb{R}}^3$ and has no global scalar potential.

Integrating the first component and checking the remaining partial derivatives givesThis explicit scalar potential proves that the field is conservative.

Parametrize the straight path by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le \pi$. Then ${\mathbf{F}}_3\cdot d\mathbf{r}=(4t^3+\sin t)dt$, so direct evaluation of the line integral givesThe fundamental theorem for line integrals supplies an independent check: this is ${\varphi }_3(\pi ,\pi ,\pi )-{\varphi }_3(0,0,0)$.

On the first segment only $z$ varies and $F_z=x^{2}y=0$. On the second only $y$ varies and $F_y=x^{2}z=0$. On the last segment $y=z=\pi$, so the line integral isThe agreement is the expected path independence of a conservative vector field.