= Solution
On the first straight segment, $y=z=0$ and only $dx$ is nonzero, so both integrals vanish. On the second segment, $x=1,z=0$ and only $dy$ is nonzero; each integrand is one, contributing $1$. On the last segment, $x=y=1$ and only $dz$ is nonzero; each integrand is $\sinh z$, contributing $\cosh1-1$. Thus
$$
\boxed{\int_C\mathbf F\cdot d\mathbf x=\cosh1,\qquad
\int_C\mathbf G\cdot d\mathbf x=\cosh1}.
$$
The second result also follows from the <fundamental theorem for line integrals>, using the potential found in (a). Agreement on this one path does not make $\mathbf F$ conservative.
Back to article page
