082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

894 Chapter 27 Line integrals and multiple integrals
=
∫ x=1
x=0
= [x 2 +x] 1 0
= 2
2x+1dx
Example27.17 Evaluate ∮ S v·dSwherev =x2 i+ 1 2 y2 j + 1 2 z2 k and whereSis the surface of the unit
cube 0 x 1, 0 y 1, 0 z 1. The vector dS should be drawn on each of the
sixfaces inan outward sense.
Solution ThecubeisshowninFigure27.22.Weevaluatethesurfaceintegralovereachofthesix
faces separately and then add the results.
OnsurfaceA,x = 1and0 y1,0 z1.dSisavectornormaltothissurface,
drawn inanoutward sense,and so wecan write itas dydzi. Then
v·dS= ( x 2 i + 1 2 y2 j+ 1 2 z2 k ) ·dydzi
=x 2 dydz
=dydz
sinceonA,x=1
The required surfaceintegral overAisthen
∫ z=1 ∫ y=1
z=0
y=0
1dydz=
= 1
∫ z=1
z=0
[y] 1 0 dz
OnsurfaceB,x = 0and0 y1,0 z1.dSisavectornormaltothissurfaceand
sowecanwriteitas −dydzi.Thenv·dSbecomes −x 2 dydz = 0sincex = 0.Overthis
surface, the integral iszero.
You should verify in a similar manner that over each ofC and E the integral is 1 2 ,
whilst integrals overDandF arezero. The total surfaceintegral isthen 2.
on E, dS = dxdy k
z
z = 1
E
dx
dy
D
B
dz
dx
on C, dS = dxdz j
x
x = 1
A
y= 1
C
dy
F
dz
on A, dS = dydz i
y
Figure27.22
Theintegral isevaluated over
allsixsurfaces.