Introduction to Vector Analysis by Harry F. Davis & Arthur David Snider

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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