Chapter 15
Section 15.2: Line Integrals

• For a curve in a plane given by the parametric `bbr(t) = x(t)bbi + y(t)bbj, where\quad a <= t <= b ` the curve is smooth if `(dx)/(dt) \quad and \quad (dy)/(dt)` are continuous on the closed inteval and where we do not have both equal to zero at the same value of `t`.
• Similarly, in space, for the curve `bbr(t)=x(t)bbi + y(t)bbj +z(t)bbk,` we look at `(dx)/(dt), \quad (dy)/(dt), \quad and \quad (dz)/(dt)` with corresponding stipulations.
• Piecewise smooth curves
• Review For `bbr(t)=f(t)bbi+g(t)bbj+h(t)bbk` we have
  `bbr^'=(dbbr)/(dt)=(df)/(dt)bbi+(dg)/(dt)bbj+(dh)/(dt)bbk`. If we consider `t` to be time then `dt` is the change in time and each of `df, dg, and dh` is a change in position, so `(dbbr)/(dt)` is the velocity vector. Thus,
  1. Velocity: `bbv=(dbbr)/(dt) \quad`
  2. Speed = `||bbv||`
  3. Acceleration: `bba = (dbbv)/(dt) = (d^2bbr)/(dt^2)`
  4. Direction of motion is `(bbv)/(||bbv||)`
• Definition of Line Integral:
  (Note that we use `Delta s` and `ds` as the length of a sub-arc of the curve.]
  If `f` is defined in a region containing a smooth curve `C` of finite length, then the line integral of `f` along `C` is given by

  `int_C f(x,y)ds = lim_(||Delta||rarr0)sum_(i=1)^nf(x_i,y_i)Delta s_i`
  or `int_C f(x,y,z)ds = lim_(||Delta||rarr0)sum_(i=1)^nf(x_i,y_i,z_i)Delta s_i`
  provided the limit exists.

• Evaluation of a line integral as a definite integral
  Let `f` be continuous in a region containing a smooth curve `C`.
  If `C` is given by `bbr(t) = x(t)bbi +y(t)bbj` where `a <= t <=b`, then

  `int_C f(x,y)ds = int_a^b f(x(t),y(t)) sqrt( [x^'(t)]^2 + [y^'(t)]^2 ) \quad dt`.

  If `C` is given by `bbr(t) = x(t)bbi +y(t)bbj +z(t)bbk` where `a <= t <=b`, then

  `int_C f(x,y,z)ds = int_a^b f(x(t),y(t),z(t)) sqrt( [x^'(t)]^2 + [y^'(t)]^2 + [z^'(t)]^2) \quad dt`.
  However, we note that `sqrt( [x^'(t)]^2 + [y^'(t)]^2 + [z^'(t)]^2)` is `||bbv(t)||` so we can rewrite the integral as
  `int_C f(x,y,z)ds = int_a^b f(x(t),y(t),z(t))\quad ||bbv(t)|| \quad dt`.

• Special case, where f(x,y)=1, or f(x,y,z)=1,
  `int_C 1 ds = int_a^b ||bbr^'(t)||dt = "length of curve " C`
  This is really the area of a one unit high ribbon sitting on top of the curve. But the numeric value of that area is exacly equal to the length of the curve.
• Compare this, `int_C 1ds = int_a^b ||bbr^'(t)||dt = int_a^b sqrt( (x^'(t))^2 +(y^'(t))^2) dt `, to the length of the arc for `y=f(x)` from `x=a` to `x=b`, which we , back in Chapter 7, Section 4, as `int_a^b sqrt( 1 + (f^'(x))^2)dx`
• Simple, straight line approach to work
  Work = force * distance
  Work = (component of force in the direction of motion)(distance moved)
  Work = (Magnitude of force applied)*cos(`theta`) )(magnitude of distance moved) where `theta` is the angle between the direction of the force and the direction of displacement
  Work = `||bbF||*cos( theta )*||displacement||`
  Work = `||bbF||*||Delta bbr||*cos( theta ) `
  Work = `bbF•dbbr`
• Sample solutions to such problems.
• Now we want to consider the direction of motion as changing (i.e., the object is moving in some complex curve) and the possibility that the force is not constant everywhere, that it is a force field, a vector field.
• In the case of a vector field, the total work done, W, is given by
  ` W = int_CbbF(x,y,z)•bbT(x,y,z)\quadds` where `bbT` is the unit tangent vector at each point.
• `bbF•bbT\quadds = bbF•(bbr^'(t))/(||bbr^'(t)||)||bbr^'(t)||dt`
  `bbF•bbT\quadds = bbF•bbr^'(t)dt`
  `bbF•bbT\quadds = bbF•dbbr`
• Definition of the line integral of a vector field:
  Let `bbf` be a continuous vector field defined on a smooth curve `C` given by `bbr(t), a <= t <= b` (and then we have `bbT` as the unit tangent vectors at each point on `bbr`, and `Deltas` a subarc of the curve, giving rise the the `ds` in the integral). The line integral of `bbF` on `C` is given by:

  `int_CbbF•dr = int_C bbF•bbT ds = int_a^b bbF(x(t),y(t),z(t))•bbr^'(t) dt`.

• The orientation of the curve `C` is important in that it gives the direction of motion so
  `int_(-C)bbF•dbbr = - int_CbbF•dbbr`
• Line integrals in Differential Form (alternative form):
  [This gets a bit confusing in that we, once again, run out of symbols. As a result of this, we use symbols whose meaning comes from the context of their use. Thus, in the following material, `dx` is meant to refer to the derivative of the `bbi` component of the `bbr(t)` function and is really `dx = bbr_x^'(t)dt` where I have used `bbr_x^'` to refer to the function giving the magnitude, in `bbr` of the `bbi` component, that is `x^'(t)`. Similarly, `dy` is really `bbr_y^'(t)dt`, that is `y^'(t)`.] If `bbF` is a vector field of the form `bbF(x,y) = Mbbi+Nbbj`, and `C` is given by `bbr(t) = x(t)bbi + y(t)bbj`, then `bbF•dr` is often written as `M dx + Ndy`.
  `int_CbbF•dr = int_CbbF•(dr)/(dt) dt`
  `int_CbbF•dr = int_a^b (Mbbi + Nbbj)• (x^'(t)bbi + y^'(t)bbj) dt`
  `int_CbbF•dr = int_a^b (M(dx)/(dt) + N(dy)/(dt)) dt`
  `int_CbbF•dr = int_a^b (M dx + N dy)`
  `int_CbbF•dr = int_C (M dx + N dy)`
  
• Extend the differential form to 3 variables:
  `int_CbbF•dr = int_C (M dx + N dy + P dz)`

Some worked problems.