Alternate notation for line integrals

We initially denoted line integrals using the notation

$\displaystyle \int_{{C}}^{}$F . ds = $\displaystyle \int_{{a}}^{{b}}$F(c(t)) . c'(t)dt (1)

where c(t) = (x(t), y(t), z(t)). We can derive a new notation by "multiplying out" the dot products on both sides of equation (1).

Let's assume we are in three dimensions to F = (F1, F2, F3). If we multiply out the left hand side, thinking of ds being the "vector" ds = (dx, dy, dz), it becomes

$\displaystyle \int_{{C}}^{}$F . ds = $\displaystyle \int_{C}^{}$F1dx + F2dy + F3dz.    

Also, since

c'(t) = $\displaystyle \left(\vphantom{ \frac{\mathrm{d} x}{\mathrm{d} t}, \frac{\mathrm{d} y}{\mathrm{d} t}, \frac{\mathrm{d} z}{\mathrm{d} t}}\right.$$\displaystyle {\frac{{\mathrm{d} x}}{{\mathrm{d} t}}}$,$\displaystyle {\frac{{\mathrm{d} y}}{{\mathrm{d} t}}}$,$\displaystyle {\frac{{\mathrm{d} z}}{{\mathrm{d} t}}}$$\displaystyle \left.\vphantom{ \frac{\mathrm{d} x}{\mathrm{d} t}, \frac{\mathrm{d} y}{\mathrm{d} t}, \frac{\mathrm{d} z}{\mathrm{d} t}}\right)$,    

we could "multiply out" the right hand side of of equation (1) to write it as

$\displaystyle \int_{{a}}^{{b}}$F(c(t)) . c'(t)dt = $\displaystyle \int_{a}^{b}$$\displaystyle \left(\vphantom{F_1(\mathbf{c}(t)) \frac{\mathrm{d} x}{\mathrm{d}... ...y}{\mathrm{d} t} + F_3(\mathbf{c}(t)) \frac{\mathrm{d} z}{\mathrm{d} t}}\right.$F1(c(t))$\displaystyle {\frac{{\mathrm{d} x}}{{\mathrm{d} t}}}$ + F2(c(t))$\displaystyle {\frac{{\mathrm{d} y}}{{\mathrm{d} t}}}$ + F3(c(t))$\displaystyle {\frac{{\mathrm{d} z}}{{\mathrm{d} t}}}$$\displaystyle \left.\vphantom{F_1(\mathbf{c}(t)) \frac{\mathrm{d} x}{\mathrm{d}... ...y}{\mathrm{d} t} + F_3(\mathbf{c}(t)) \frac{\mathrm{d} z}{\mathrm{d} t}}\right)$dt.    

Frequently, line integral problems are presented in this notation. In this case, one can compute them directly using the formula

$\displaystyle \int_{C}^{}$F1dx + F2dy + F3dz = $\displaystyle \int_{a}^{b}$$\displaystyle \left(\vphantom{F_1(\mathbf{c}(t)) \frac{\mathrm{d} x}{\mathrm{d}... ...y}{\mathrm{d} t} + F_3(\mathbf{c}(t)) \frac{\mathrm{d} z}{\mathrm{d} t}}\right.$F1(c(t))$\displaystyle {\frac{{\mathrm{d} x}}{{\mathrm{d} t}}}$ + F2(c(t))$\displaystyle {\frac{{\mathrm{d} y}}{{\mathrm{d} t}}}$ + F3(c(t))$\displaystyle {\frac{{\mathrm{d} z}}{{\mathrm{d} t}}}$$\displaystyle \left.\vphantom{F_1(\mathbf{c}(t)) \frac{\mathrm{d} x}{\mathrm{d}... ...y}{\mathrm{d} t} + F_3(\mathbf{c}(t)) \frac{\mathrm{d} z}{\mathrm{d} t}}\right)$dt. (2)

Example

Evaluate

$\displaystyle \int_{C}^{}$y dx + (x + y)dy + dz    

where thee curve C is parameterized by c(t) = (t, 1 - t, 1), 0$ \le$t$ \le$1.

Solution: If we write c(t) = (t, 1 - t, 1) = (x(t), y(t), z(t)), Then $ {\frac{{\mathrm{d} x}}{{\mathrm{d} t}}}$ = 1, $ {\frac{{\mathrm{d} y}}{{\mathrm{d} t}}}$ = - 1, and $ {\frac{{\mathrm{d} z}}{{\mathrm{d} t}}}$ = 0. Since F1(x, y, z) = y, F2(x, y, z) = x + y, and F3(x, y, z) = 1, we can use formula (2) to calculate the integral.

  $\displaystyle \int_{C}^{}$y dx + (x + y)dy    
  = $\displaystyle \int_{0}^{1}$$\displaystyle \left(\vphantom{y(t) \frac{\mathrm{d} x}{\mathrm{d} t} + (x(t)+y(t))\frac{\mathrm{d} y}{\mathrm{d} t} + \frac{\mathrm{d} z}{\mathrm{d} t} }\right.$y(t)$\displaystyle {\frac{{\mathrm{d} x}}{{\mathrm{d} t}}}$ + (x(t) + y(t))$\displaystyle {\frac{{\mathrm{d} y}}{{\mathrm{d} t}}}$ + $\displaystyle {\frac{{\mathrm{d} z}}{{\mathrm{d} t}}}$$\displaystyle \left.\vphantom{y(t) \frac{\mathrm{d} x}{\mathrm{d} t} + (x(t)+y(t))\frac{\mathrm{d} y}{\mathrm{d} t} + \frac{\mathrm{d} z}{\mathrm{d} t} }\right)$dt    
  = $\displaystyle \int_{0}^{1}$$\displaystyle \left[\vphantom{(1-t)(1) + (t + (1-t)) (-1)}\right.$(1 - t)(1) + (t + (1 - t))(- 1)$\displaystyle \left.\vphantom{(1-t)(1) + (t + (1-t)) (-1)}\right]$dt    
  = $\displaystyle \int_{0}^{1}$[(1 - t) + (- 1)]dt    
  = $\displaystyle \int_{0}^{1}$(- t)dt = - $\displaystyle {\frac{{t^2}}{{2}}}$$\displaystyle \Big\vert _{0}^{1}$ = - $\displaystyle {\frac{{1}}{{2}}}$    


Duane Nykamp
nykamp@math.umn.edu
2005-10-27