Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 5

Parametrize the straight line by $r(t)=(t,t,t)$, $0\le t\le 1$, with $r^{\prime }(t)=(1,1,1)$. Substituting into the vector field and taking the dot product givesHence the direct line integral is

For the curved path $r(t)=(t,t^2,t^3)$, $r^{\prime }(t)=(1,2t,3t^2)$. The vector field along this parametric curve isIts dot product with the tangent is $(t^2+2t)e^t+13t^{12}$. Therefore the direct line integral is

Compute the curl in Cartesian coordinates:The three components are $6y{z}^2-6y{z}^2$, $0-0$, and $e^x-e^x$. Consequently

Integrate $f_x=y{e}^x$ with respect to $x$, obtaining $f=y{e}^x+g(y,z)$. Matching $f_y=e^x+2y{z}^3$ gives $g_y=2y{z}^3$, so $g=y^2{z}^3+h(z)$. Finally, $f_z=3y^2{z}^2+h^{\prime }(z)$ must equal the third component, so $h^{\prime }=0$. Thus a scalar potential isThe fundamental theorem for line integrals now gives ${\int }_{C}v\cdot dr=f(1,1,1)-f(0,0,0)=e+1$ for either path, confirming both direct computations. Since the curl of a gradient is zero for this smooth function, it also confirms the result in part (b). The potential is defined throughout ${\mathbb{R}}^3$, so no domain obstruction to path independence occurs.