= Solution
For $\mathbf r(t)=(t,t,t^2)$, the tangent <vector> is $\mathbf r'(t)=(1,1,2t)$. Substitute into the field:
$$
\mathbf F(\mathbf r(t))=(2t^2-t^6,\ t^2-2t,\ -3t^5-1).
$$
Its <scalar product> with the tangent is $3t^2-4t-7t^6$. Hence the explicitly parametrized <line integral> is
$$
\boxed{\int_C\mathbf F\cdot d\mathbf r
=\int_1^2(3t^2-4t-7t^6)\,dt
=[t^3-2t^2-t^7]_1^2=-126}.
$$
As a check, the potential from part (d) has values $-128+C$ at $(2,2,4)$ and $-2+C$ at $(1,1,1)$, giving the same difference.
Back to article page
