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We can use the Taylor series to calculate numerical approximations to
integrals too. Let's look at integrating the function f (x) from x_{0}
to x_{0} + h numerically. Expanding f (x) around the midpoint of the
interval we have
f (x)dx

=

x
 

=

f (a + h/2)h + O(h^{ 3})
 (13) 
To apply this to the integral from a to b , we can divide it into
N subintervals with
h

=

 
x_{i}

=

a + h ,
 (14) 
and
dxf (x)

=

f (x_{i})h + N O(h^{ 3})
 

=

f (x_{i}) + O(N^{  2})
 (15) 
It is interesting to notice that round off will produce an error
of
approximately rf (x) in f (x) , where r is the precision defined in
section 1, which bounded by
(b  a)rmax (f (x)) , which is independent
of N if the sum is done carefully. Integrations tend to be much more
stable than derivatives when decreasing the step size.
Now that we know how to do a simple integral let's see how to do
the various kinds of integrals we will see in electrodynamics.
Often we need line integrals like
either around a loop or along a path between two end points.
One way is to first define the path parametrically.
For example if we need to calculate the line integral of
()
around
an ellipse with semimajor axis of length a parallel to , semiminor
axis of length b in the y  z plane making an angle of with
the y axis, we can define the path parametrically as
(u) = acos(u) + bsin(u)cos() + bsin(u)sin()


 (17) 
where u runs from 0 to 2 .
The line integral can be approximated by
u_{i}


i
 


 


(u_{i})  (u_{i  1})
 
(x,y,z) d

=

() + O(N^{  2})
 (18) 
Notice that this is written in a form exactly analogous to Eq. 15,
and is an explicit example of what is shown in figure 1.3 of Jackson.
The generalization to other paths should now be easy.
Next we need to consider twodimensional integrals. One way to perform
these is to apply the onedimensional method for each dimension, or
equivalently Taylor series expand in both x and y . To integrate
over a small rectangle of sides h_{a} and h_{b} ,
f (x,y)dxdy

=

 

+

f_{x}(a + h_{a}/2,b + h_{b}/2)(x  a  )
 

+

xdy
 

=

f (a + h_{a}/2,b + h_{b}/2)h_{a}h_{b} + O(h^{ 4})
 (19) 
where O(h^{ 4}) etc. means that order of products of the
factors h_{a} and h_{b} , and f_{x} and f_{y} are the partial derivatives
of f with respect to x and y . Notice this is nothing but the value
of the function at the center of the rectangle multiplied by the
area of the rectangle. If we integrate over an area using N points
total (that is N little rectangles or roughly N^{ 1/2} in each direction),
the integration error will be O(N^{  1}) . Notice we can easily generalize
this to other basic shapes by multiplying a value of the function
at a central point times the area of the shape.
Going through the same mathematics for a three dimensional integral,
we get the obvious generalization of the value of the function at
the center of the 3d rectangular prism times the volume of
the prism,

f (x,y,z)dxdydz = f (a + h_{a}/2,b + h_{b}/2,c + h_{c}/2)h_{a}h_{b}h_{c} + O(h^{ 5})
 (20)

and if we integrate over a volume using N points total, the integration
error will be O(N^{  2/3}) . For a general d dimensional integration,
the midpoint rule will give an error that goes like O(N^{  2/d}) .
Again we can generalize to other shapes by multiplying a value of the function
at a central point times the volume of the shape.
It is instructive to do the line integral of around the outside
of a very small square of
side h centered at
(x_{0},y_{0},z_{0}) .
Let's take the square to be oriented along the x and y axes.
The line integral is,

() d = h[E_{y}(x_{0} + h/2,y_{0},z_{0})  E_{x}(x_{0},y_{0} + h/2,z_{0})  E_{y}(x_{0}  h/2,y_{0},z0) + E_{x}(x_{0},y_{0}  h/2,z0)] .
 (21)

Notice that the right hand side is our central difference
formula for the z component of the curl at the center of the
square, multiplied by h^{ 2}. That is, we have proved Stokes theorem.
A similar application to a surface integral over a little cube will
give the divergence theorem.
Let's see how to apply our integration to a surface integral like those
typically found in electromagnetism. Often we want to integrate the
surface normal component of a vector field over a surface. For example,
we might want to integrate the normal component of the electric field
over a surface
Rather than do a trivial problem like the surface of
a cube or sphere, lets integrate over an ellipsoidal surface.
We do this problem in two ways. First we imagine we are going to set up
the problem analytically and then do the double integral numerically
at the end. Second we will imagine doing it numerically from the beginning.
One way to
define the surface is parametrically, similarly to the way we defined
a path parametrically. We pick cartesian coordinates so that the
origin is at the center of our ellipsoid with the coordinate
axes aligned with the semiaxes of the ellipsoid. The surface can be
given by
x(u,v)

=

asin(u)cos (v)
 
y(u,v)

=

bsin(u)sin (v)
 
z(u,v)

=

ccos (u)
 (23) 
with
0 u ,
0 v 2 . As with any parametric
definition, there are an infinity of choices you can make.
This parameterization defines the vector

(u,v) = x(u,v) + y(u,v) + z(u,v)
 (24)

which describes our surface. The directed area of the surface corresponding to
a small increment in u and v is

d = udv ,
 (25)

where we have chosen the sign for our particular problem so that d
points outward.
The analytic
expression for the surface integral is

dvdu (x(u,v),y(u,v),z(u,v))
 (26)

Note: if you have trouble following this to here, you are having trouble
with standard analytic vector calculus not numerical vector calculus.
To attempt an analytic integration, you would do the derivatives above
and simplify the expression.
We will now convert our expression into a form ready for
numerical calculation. We could do
the derivatives analytically, but for amusement lets do the
integrals using the midpoint rule and the derivatives using the
central differences.
h_{u}

=

 
h_{v}

=

 
u_{i}

=

(i  )h_{u}
 
v_{j}

=

(j  )h_{v}
 

=

 
d

=

(x(u_{i},v_{j}),y(u_{i},v_{j}),z(u_{i},v_{j}))
 (27) 
Alternatively, you can think about doing this surface integral
from a numerical point of view from the beginning. In that case,
we could break up u and v into equally spaced intervals. This
would divide the surface into little 4 sided areas (actually at
u = 0 and u = one of the sides goes to zero and they become 3 sided
there, but this can be ignored for now) and
we can think of doing the integration by multiplying the surface
area made up of the little areas bounded by the four surface points

= (u_{i} h_{u}/2,v_{j} h_{v}/2)
 (28)

If we approximate this area by that of a straight sided quadrilateral
with an outward direction we get
the cross products of the little vectors that go across the
quadrilateral or an equivalent expression.
This is just the cross product given in Eq. 27
above. Notice that we can use this ``numerical'' thinking to derive
or check our analytic results and gain a better understanding of
how everything fits together.
Next: 3 Conclusion
Up: Introduction to some Basic
Previous: 1 Numerical Derivatives