= Solution
Parametrize the diagonal by $\mathbf x(t)=(t,t,t)$ for $0\leq t\leq1$. Then
$$
\mathbf F\cdot\mathbf x'=2t^2\cosh t+t^3\sinh t,
\qquad \mathbf G\cdot\mathbf x'=3t^2\cosh t+t^3\sinh t.
$$
The second expression is $(t^3\cosh t)'$, immediately giving $\int\mathbf G\cdot d\mathbf x=\cosh1$. The first differs from it by $t^2\cosh t$. Two applications of <integration by parts> give
$$
\int_0^1t^2\cosh t\,dt
=\left[t^2\sinh t-2t\cosh t+2\sinh t\right]_0^1
=3\sinh1-2\cosh1.
$$
Hence
$$
\boxed{\int_C\mathbf F\cdot d\mathbf x=3(\cosh1-\sinh1)=\frac3e},\qquad
\boxed{\int_C\mathbf G\cdot d\mathbf x=\cosh1}.
$$
The changed value for $\mathbf F$ explicitly demonstrates path dependence, while $\mathbf G$ gives the same endpoint-dependent result.
Back to article page
