Line Integrals

Definition of a Line Integral

By this time you should be used to the construction of an integral.  We break a geometrical figure into tiny pieces, multiply the size of the piece by the function value on that piece and add up all the products.  For one variable integration the geometrical figure is a line segment, for double integration the figure is a region, and for triple integration the figure is a solid. 

The geometrical figure of the day will be a curve.  If we have a function defined on a curve we can break up the curve into tiny line segments, multiply the length of the line segments by the function value on the segment and add up all the products.  As always, we will take a limit as the length of the line segments approaches zero.  This new quantity is called the line integral and can be defined in two, three, or higher dimensions.

Suppose that a wire has as density f(x,y,z) at the point (x,y,z) on the wire.  Then the line integral will equal the total mass of the wire.  Below is the definition in symbols.

Definition of the Line Integral

Let f be a function defined on a curve C of finite length.  Then the line integral of f along C is 

                  (for two dimensions)

            (for three dimensions)


Evaluating Line Integrals

This definition is not very useful by itself for finding exact line integrals.  If data is provided, then we can use it as a guide for an approximate answer.  Fortunately, there is an easier way to find the line integral when the curve is given parametrically or as a vector valued function.  We will explain how this is done for curves in R2.  The case for R3 is similar.  


        r(t)  =  x(t)i + y(t)j

be a differentiable vector valued function.  Then


We are now ready to state the theorem that shows us how to compute a line integral.


Theorem:  Line Integrals Over Vector Valued Functions


               r(t)  =  x(t)i + y(t)j          a < t < b

be a differentiable vector valued function that defines a smooth curve C.  Then


and for three dimensions, if

                r(t)  =  x(t)i + y(t)j + z(t)k          a < t < b





Find the line integral


where C is the ellipse 

        r(t)  =  (2cos t)i + (3sin t)j          0  <  t  <  2p

You may use a calculator or computer to evaluate the final integral.


We find


We have the integral


With the help of a machine, we get  




The main application of line integrals is finding the work done on an object in a force field.  If an object is moving along a curve through a force field F, then we can calculate the total work done by the force field by cutting the curve up into tiny pieces.  The work done W along each piece will be approximately equal to 

        dW  =  F . Tds

Now recall that 

        T  =              

and that 

        ds  =  ||r'(t)||dt


        dW  =  F . r'(t) dt

As usual, we add up all the small pieces of work and take the limit as the pieces get small to end up with an integral.

Definition of Work

Let F be a vector field and C be a curve defined by the vector valued function r.  Then the work done by F on an object moving along C is given by 




Find the work done by the vector field

        F(x,y,z)  =  xi + 3xyj - (x + z)k

on a particle moving along the line segment that goes from (1,4,2) to (0,5,1)



We first have to parameterize the curve.  We have 

        r(t)  =  <1,4,2> + [<0,5,1> - <1,4,2>]t  =  <1 - t, 4 + t, 2 - t>


        r'(t)  =  -i + j - k

Taking the dot product, we get

        F . r'(t)  =  -x + 3xy + x + z  =  3xy + z

        =  3(1 - t)(4 + t) + (2 - t)  =  -3t2 -10t + 14

Now we just integrate


Notice that work done by a force field on an object moving along a curve depends on the direction that the object goes.  In fact the opposite direction will produce the negative of the work done in the original direction.  This is clear from the fact that everything is the same except the order which we write a and b.

Line Integrals in Differential Form

We can rewrite r'(t)dt as

          dr                 dx          dy          dz
                 dt  =  (          i+          j +          k ) dt 
           dt                 dt           dt           dt

              =  dx i + dyj + dzk 

So that if 

        F  =  Mi + Nj + Pk


        F . r'(t)dt  =  Mdx + Ndy + Pdz

This is called the differential form.





where C is the part of the helix

        r(t)  =  sin t i + cos t j + t k        <  t  <  2p



We have 

        r'(t)  =  cos t i - sin t j + k  

so that 

        ydx + zdy  =  (cos2t - t sin t)dt

This leads us to the integral


with a little bit of effort (using integration by parts) we get



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