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## Integration of a vector field along a curve

Given a curve $C.$ in space with starting point A and end point B in parametric form

$x=x(t),y=y(t),z=z(t),tA.≤t≤tB..$

This curve can be a vector

$r(t)=x(t)e1+y(t)e2+z(t)e3$

assign whose peak the curve when varying the parameter $t$ from $tA.$ after $tB.$ passes through.

The curve is differentiable if $x(t)$, $y(t)$ and $z(t)$ differentiable functions of $t$ are, and then has the tangent vector

$dr(t)dt=r'(t)=x'(t)e1+y'(t)e2+z'(t)e3 .$

The curve piece $dr=r'(t)dt$ is tangent to the curve at that point $r(t)$.

Well be

$F.(x,y,z)=F.1(x,y,z)e1+F.2(x,y,z)e2+F.3(x,y,z)e3$

a vector field. We define the line integral of the vector field $F.$ along the curve $C.$ as

$I.=∫C.F.⋅dr=∫tA.tB.F.(x(t),y(t),z(t))⋅r'(t)dt,$

i.e., $I.$ is the integral of the scalar product of $F.$ with $dr$. Here is the scalar product

$F.(x(t),y(t),z(t))⋅r'(t)=F.1x'(t)+F.2y'(t)+F.3z'(t).$

### Calculation of work

As an example, consider the work involved in moving a body in a force field $F.(x,y,z)$ is performed. If the body moves a distance $dr$, will be a job

$dW.=F.⋅dr$

performed. You move the body along a curve $C.$ from point A to point B and adds up the individual distances $dr$to get all the work done

$W.=∫C.F.⋅dr.$

If there is a conservative force field, it can be described as a gradient of a scalar field $φ(x,y,z)$ grasp

$F.=-∇φ=-∂φ∂xe1+∂φ∂ye2+∂φ∂ze3.$

The work done is then

$W.=-∫C.φxdx+φydy+φzdz=φ(xA.,yA.,zA.)-φ(xB.,yB.,zB.),$

i.e., with a conservative force field, the work done is path-independent. In particular, the work for a closed curve is zero.

Examples of conservative force fields are the electric field and the gravitational field. A non-conservative vector field is the magnetic field.