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 atba \leq t \leq b, and let ff be a continuous scalar function on CC. The scalar line integral of ff along CC is Cfds=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), atba \leq t \leq b. The vector line integral (or work integral) of F\mathbf{F} along CC is CFdr=abF(r(t))r(t)dt=CPdx+Qdy+Rdz\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: CFdr=CFdr\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)=(cost,sint,t)\mathbf{r}(t) = (\cos t, \sin t, t) for 0t2π0 \leq t \leq 2\pi under the force F=(y,x,1)\mathbf{F} = (y, -x, 1): r(t)=(sint,cost,1)\mathbf{r}'(t) = (-\sin t, \cos t, 1) CFdr=02π(sint)(sint)+(cost)(cost)+1dt=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: (sint)(sint)+(cost)(cost)+1=sin2tcos2t+1=0(\sin t)(-\sin t) + (-\cos t)(\cos t) + 1 = -\sin^2 t - \cos^2 t + 1 = 0. So CFdr=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)=(cost,sint)\mathbf{r}(t) = (\cos t, \sin t), 0t2π0 \leq t \leq 2\pi: CFdr=02π[(sint)(sint)+(cost)(cost)]dt=02π1dt=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 Cfds\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 CFdr\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).