Line Integrals (continued)

The line integral can also be expressed in terms of each coordinate variable (x, y, z).

Suppose F(x, y, z) is a vector field defined as

F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k

where P(x, y, z), Q(x, y, z), and R(x, y, z) are continuous functions of x, y, and z

and dr is a vector element of length along the curve C in space

here dr = dx i + dy j + dz k Then the dot product

F(x, y, z) · dr yields the following line integral along the curve C in space

∫ P(x, y, z) dx + ∫ Q(x, y, z) dy + ∫ R(x, y, z) dz

C C C

Both dot products discussed here F(x, y) · dr(x, y) and F(x, y¸z) · dr(x, y,z)

appear in engineering as the incremental work of the “force” F along the path, C. i.e.

Suppose F is a force acting on a particle in the x-y plane and r is the position vector

from the origin to the particle.

Let the particle move an amount ds along the curve, C, in the x-y plane under the

influence of the force F . The change in the position vector along the curve

(tangent to C) is dr . Then the incremental work, dW, done on the particle by the

force F is the dot product of F and dr . Thus the line integral along C gives the

work, W, done by the force F acting on the particle as it moves along C.

W = ∫ F . dr = ∫ F . (dr/dt) dt and dr = T ds, T is the unit tangential vector to the curve, C

C

Click here for further discussion of line integrals.

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