Introduction to Vector Analysis by Harry F. Davis & Arthur David Snider
Author:Harry F. Davis & Arthur David Snider
Language: eng
Format: epub
Tags: Mathematics; Vector Algebra; Vector and Scalar Fields; Line and Surface Integrals; Generalised Orthogonal Coordinates; Vector Equations of Electromagnetism
(u + Au, v + Av)
u + Au, v)
FIGURE 4.15
Line and Surface Integrals
CHAP. 4
the further approximations
dR
R(u + Aw, v) — R(u,v) &~Au du
R(u, v + Av) — R(u,v)
dR
dv
Av
we find that the area of the patch is given approximately by
AS
dR dR
— x —
du dv
AuAv
Summing these up over the surface and letting Au and Ay go to zero, we argue that the surface area of a regular surface element is given by
If we introduce the notation
dR dR
du dv
dudv
,o SR 1 dR ,
aS = — du x — dv
du dv
(4.25)
(4.26)
then we see that dS is a vector normal to the surface at P, whose magnitude dS = |dS| is the element of area. The integral (4.26) may be written in the alternative forms
or
or even
tt ds
JJn • dS
where n is a unit normal in the same direction as dS.
Example 4.12 Find the surface area of the surface defined by the equations
x = cos u y = sin u z = v for 0 < u < In, 0 < v < 1. Solution We have
dR
du dR
dv
dS
— sin u i + cos u j = k
i J k
— sin u cos u 0 0 0 1
du dv = (cos u i + sin u j) du dv \dS\ — I J (cos 2 u + sin 2 uf dudv = 2%
SEC. 4.6
Oriented Surfaces
161
This surface is a right circular cylinder of unit radius and unit height. If it is cut along a "seam", it unfolds into a rectangle whose dimensions are 2% by 1.
We now consider a special case of Eq. (4.25) that will illustrate further its geometrical significance. Let us suppose that the surface we consider is given in the form z = f(x,y). In other words, we are told how far above the xy plane the surface is for each point (x,y) in the xy plane. Then it is convenient to use x and y instead of u and v as the parameters. Let us suppose that the projection of the surface element on the xy plane is bounded by the curves
y = yi(x) y = y 2 (x) x = a x = b
as shown in Fig. 4.16.
We can write x — u, y = v, and z = f{u,v), in order to make use of the preceding formulas. We have
aR_<3R_. fd£\ du dx \dx)
dR dR
dv dy
Taking the vector cross product,
i j k
'♦'£>
dR dR
du dv
i o V
OX
0 1
df
dy
= k
dx dy
The magnitude of this vector is ^/l + (df/dx) 2 + (df/dy) 2 , so that the integral (4.25) is
rb ry 2 (x)
Ja Jyi(x)
w-iIHf) 2 ^
(4.27)
The geometrical significance of this is seen by considering the angle y between dS and k. By a simple calculation using scalar products we see that
|cos y\ = so that (4.27) is simply
\dS • k| |<2S|
-SHf
b Cy 2 (x) dx dy (*) Icos y\
Ja Jyi
(4.28)
This integral could have been obtained heuristically by considering the area cosine principle which says that, if we look at a plane area A whose normal makes an acute angle 6 with the line of
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