ConceptComplete

Line Integrals

Line integrals generalize ordinary integrals to integration along curves, computing quantities such as work done by a force field along a path or the mass of a wire with varying density.


Scalar Line Integrals

Definition

Let CC be a smooth curve parametrized by r(t)=(x(t),y(t),z(t))\mathbf{r}(t) = (x(t), y(t), z(t)) for a≤t≤ba \leq t \leq b, and let ff be a continuous scalar function on CC. The scalar line integral of ff along CC is ∫Cf ds=∫abf(r(t))∥r′(t)∥ dt\int_C f\,ds = \int_a^b f(\mathbf{r}(t)) \|\mathbf{r}'(t)\|\,dt where ds=∥r′(t)∥ dtds = \|\mathbf{r}'(t)\|\,dt is the arc length element. This integral is independent of the parametrization (and its orientation).


Vector Line Integrals

Definition

Let F\mathbf{F} be a continuous vector field and CC an oriented smooth curve parametrized by r(t)\mathbf{r}(t), a≤t≤ba \leq t \leq b. The vector line integral (or work integral) of F\mathbf{F} along CC is ∫CF⋅dr=∫abF(r(t))⋅r′(t) dt=∫CP dx+Q dy+R dz\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt = \int_C P\,dx + Q\,dy + R\,dz This measures the work done by F\mathbf{F} on a particle moving along CC. Reversing orientation changes the sign: ∫−CF⋅dr=−∫CF⋅dr\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}.


Examples

ExampleWork along a helix

The helix r(t)=(cos⁡t,sin⁡t,t)\mathbf{r}(t) = (\cos t, \sin t, t) for 0≤t≤2π0 \leq t \leq 2\pi under the force F=(y,−x,1)\mathbf{F} = (y, -x, 1): r′(t)=(−sin⁡t,cos⁡t,1)\mathbf{r}'(t) = (-\sin t, \cos t, 1) ∫CF⋅dr=∫02π(sin⁡t)(−sin⁡t)+(−cos⁡t)(cos⁡t)+1 dt=∫02π(−1+1) dt=0\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} (\sin t)(-\sin t) + (-\cos t)(\cos t) + 1\,dt = \int_0^{2\pi} (-1 + 1)\,dt = 0

Wait: (sin⁡t)(−sin⁡t)+(−cos⁡t)(cos⁡t)+1=−sin⁡2t−cos⁡2t+1=0(\sin t)(-\sin t) + (-\cos t)(\cos t) + 1 = -\sin^2 t - \cos^2 t + 1 = 0. So ∫CF⋅dr=0\int_C \mathbf{F} \cdot d\mathbf{r} = 0.

ExampleCirculation around a circle

For F=(−y,x)\mathbf{F} = (-y, x) and CC the unit circle r(t)=(cos⁡t,sin⁡t)\mathbf{r}(t) = (\cos t, \sin t), 0≤t≤2π0 \leq t \leq 2\pi: ∫CF⋅dr=∫02π[(−sin⁡t)(−sin⁡t)+(cos⁡t)(cos⁡t)] dt=∫02π1 dt=2π\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} [(-\sin t)(-\sin t) + (\cos t)(\cos t)]\,dt = \int_0^{2\pi} 1\,dt = 2\pi


RemarkPhysical interpretation

The scalar line integral ∫Cf ds\int_C f\,ds computes the total mass of a wire with linear density ff, or the total value of ff accumulated along CC. The vector line integral ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} computes the work done by the force F\mathbf{F} (the component of F\mathbf{F} tangent to CC, integrated over arc length).