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

27.7 Some simple volume and surface integrals 893
Engineeringapplication27.7
ElectricfluxandGauss’slaw
Electric charges produce an electric field, E, which can be visualized by drawing
lines of force. Suppose we surround a region containing charges with a surface S.
ThefluxofEthroughSisameasureofthenumberoflinesofforcepassingthrough
it. The flux isgiven by
∫
flux = E·dS
S
Gauss’slawstatesthatthetotalfluxoutofanyclosedsurfaceisproportionaltothe
totalchargeenclosed.Soconsideraclosedsurfaceinfreespace,enclosingavolume
V. The total charge enclosed
∫
inV isgiven by the volume integral
total charge = ρdV
Gauss’s lawthen states:
∮
E·dS= 1 ∫
ρdV
ε 0
S
V
V
where ε 0
isthepermittivityoffreespace.Thenotation ∮ indicatesthatthesurfaceis
a closed surface. Note how a surface integral can be expressed as a volume integral.
This relationship isgeneralized inthe divergence theorem inthe following section.
Example27.16 Avectorfield isgiven by v =x 2 i + 1 2 y2 j+ 1 2 z2 k.
(a) Find divv.
(b) Evaluate ∫ V divvdVwhereVistheunitcube0x1,0y1,0z1.
Solution (a) divv = 2x +y+z.
(b) We seek ∫ (2x +y+z)dV whereV is the given unit cube. The small element of
V
volume dV is equal to dzdydx. With appropriate limits of integration the integral
becomes
∫ x=1 ∫ y=1 ∫ z=1
x=0
y=0
z=0
(2x+y+z)dzdydx
This isatripleintegral of the kind evaluated inSection 27.6.2:
∫ x=1 ∫ y=1 ∫ z=1
∫ x=1 ∫ y=1
] 1
(2x+y+z)dzdydx =
[2xz+yz+ z2
dydx
x=0 y=0 z=0
x=0 y=0 2
0
∫ x=1 ∫ y=1
(
= 2x+y+ 1 )
dydx
2
=
x=0
∫ x=1
x=0
y=0
[2xy + y2
2 2] + y 1
dx
0