 Solution

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.