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The line integral ∫_C F · dr integrates a vector field along a curve. Parametrically: ∫_C F · dr = ∫_ab F(r(t)) · r'(t) dt. This represents work done by force F along the path.
A line integral of a vector field asks: what is the cumulative effect of the field along a given path? The key idea is that at each point on the path, only the component of the field in the direction of travel matters. You are essentially asking how much the field "goes along with" the motion at each step, then adding it all up.
The formal setup is: parametrize the curve C by r(t) = ⟨x(t), y(t), z(t)⟩ for t ∈ [a, b]. The tangent vector r'(t) points in the direction of travel and has magnitude equal to the speed. The line integral is then ∫_C F · dr = ∫_ab F(r(t)) · r'(t) dt. The dot product F(r(t)) · r'(t) extracts the component of F along the curve at each t — contributions where F aligns with the path are positive, where F opposes the path are negative, and where F is perpendicular the contribution is zero.
The most important physical interpretation is work. If F is a force field and a particle travels along C, the work done by F is exactly ∫_C F · dr. This makes intuitive sense: when you push an object in the direction it is already moving, you do positive work; when you push against its motion, you do negative work; when you push perpendicular to motion (like a normal force), you do no work. The dot product captures all three cases simultaneously.
Orientation matters crucially. The line integral is sensitive to which direction you traverse the path: ∫_{−C} F · dr = −∫_C F · dr, where −C is the same curve traversed backward. This is the opposite of line integrals of scalar functions (which are orientation-independent). It also sets up one of the most important concepts to come — conservative fields, where ∫_C F · dr depends only on the endpoints, not the path taken.