Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 2 / 15

Integrating the first component in $x$ gives $\varphi =x^{3}y+h(y,z)$. Matching the second requires $h_y=-2y{z}^2$, so $h=-y^2{z}^2+k(z)$. Matching the third gives $k^{\prime }=0$. Thus the field is conservative, with scalar potentialDirect differentiation verifies ${\mathbf{F}}_1=\nabla \varphi$ on all of ${\mathbb{R}}^3$.

The curl is $\nabla \times {\mathbf{F}}_2=(2z,3z^2,0)$, which is not identically zero. A gradient of a smooth scalar potential has zero curl, so ${\mathbf{F}}_2$ is not conservative on ${\mathbb{R}}^3$.

On the first segment, $x=z=0$, so both line integrals vanish. On the second, $y=1,z=0$, giving contributions ${\int }_0^{2}3x^{2}dx=8$ and ${\int }_0^{2}2xdx=4$. On the last, $x=2,y=1$, the contributions are ${\int }_0^3-2zdz=-9$ and ${\int }_0^3-6z^{2}dz=-54$. ThereforeThe first also equals $\varphi (B)-\varphi (A)=8-9=-1$, as the fundamental theorem for line integrals predicts.

The parametric curve runs from $t=0$ to $t=1$, with ${\mathbf{r}}^{\prime }=(2,3t^2,6t)$. Substitution into the two line integrals givesHenceThe first is unchanged because the field is conservative. The two different answers for the second exhibit its path dependence, consistent with its nonzero curl.