Introduction to Line Integrals

Date: 02/08/2004 at 00:31:54
From: Mike
Subject: Line integrals

Dr. Math,

I don't understand what line integrals are.  It's never made clear in
my textbook.  I have a problem where its C int(x dy + y dx).  The
problem says that:

C is the broken path line that goes from (1,1,2) to (2,0,-1) by 
following a line parallel to the x axis from (1,1,2) to (2,1,2) then 
parallel to the y axis from (2,1,2) to (2,0,2) then parallel to the z
axis from (2,0,2) to (2,0,-1).

What does all this mean?

Date: 02/08/2004 at 08:39:55
From: Doctor Jerry
Subject: Re: Line integrals

Hello Mike,

I can't undertake teaching you about line integrals, other than to say
that

int_C [ f(x,y,z)*dx + g(x,y,z)*dy + h(x,y,z)*dz ]

is evaluated by finding a parameterization 

x = p(t)
y = q(t),
z = r(t),

a <= t <= b,

for C and then

int_C [ f(x,y,z)*dx + g(x,y,z)*dy + h(x,y,z)*dz ] = 

         int(t=a,t=b,[ f(p(t),q(t),r(t))*p'(t) +
g(p(t),q(t),r(t))*q'(t)]*dt + h(p(t),q(t),r(t))*r'(t)]*dt)

If the curve is in pieces then one just parameterizes each piece.

On a piece parallel to the x-axis dy = dz = 0 and things simplify.

On the first part of the curve you have been given, from (1,1,2) to
(2,0,-1), use

x = p(t) = 1 + (2-1)t
y = q(t) = 1 + (0-1)t
z = r(t) = 2 + (-1-2)t,

t in [0,1].

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 02/08/2004 at 09:49:01
From: Mike
Subject: Thank you (Line integrals)

Thanks for your help.  It's a little clearer now.  However, I'm
wondering if you could help me understand what exactly a line integral
is.  I know that integrals are areas under a given curve but what are
line integrals?  Thank you again.

Date: 02/08/2004 at 10:44:11
From: Doctor Jerry
Subject: Re: Thank you (Line integrals)

Hello Mike,

There are several interpretations.  One of the most easily grasped 
is the following.  If we have a curve C and a vector force F(x,y,z) at
each point (x,y,z) of space, where

  F(x,y,z) = i*f(x,y,z) + j*g(x,y,z) + k*h(x,y,z),

and an object moves from one end of C to the other, then the work dW
done by F as the object moves from (x,y,z) on C a distance of ds is
approximately

  dW = F(x,y,z).T(x,y,z) * ds,

where . means dot product, T(x,y,z) is the tangent vector to the curve
C at (x,y,z), and ds is the arc length.  Because (now thinking of the
curve as parameterized with a parameter t),

  T(x,y,z) = [1/sqrt(x'(t)^2 + y'(t)^2 + z'(t)^2)] (x'(t)*i + y'(t)*j
             + z'(t)*k )

and 

  ds/dt = sqrt(x'(t)^2 + y'(t)^2 + z'(t)^2), 

we see that (now letting x'(t) = dx/dt, etc)

  dW = F(x,y,z).T(x,y,z) * ds = f(x,y,z)*dx + g(x,y,z)*dy +
       h(x,y,z)*dz.

Adding these up on the entire curve we see that 

  W = int_C [ f(x,y,z)*dx + g(x,y,z)*dy + h(x,y,z)*dz ]

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/