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

27.9 Maxwell’s equations in integral form 899
Similarly, onD, the left-hand face, dS = −dxdzj. Hence curlv·dS = dxdz. The
required integral overDis
∫ z=2 ∫ x=2
z=0
x=0
dxdz=4
On surfaceF, the base,z = 0 and dS = −dxdyk. Hence curlv·dS = dxdy. The
required integral overF is
∫ y=2 ∫ x=2
y=0
x=0
dxdy=4
RecallthatthetopsurfaceE isopen,andsowehavecompletedthesurfaceintegrals.
Finally, putting these results together,
∫
curlv·dS=−4+4−4+4+4=4
S
Notefrompart(a)thatthisequals ∮ v·drandsowehaveverifiedStokes’theorem.
EXERCISES27.8
1 VerifyStokes’theorem forthe field v =xyi +yzj
whereSisthe surface ofthe cube 0 x1,
0y1,0z1andthefacez=0isopen.
2 Ifv = 3x 2 i +xyzj −5zkverify Stokes’theorem
whereSisthe surface ofthe cube 0 x1,
0y1,0z1andthefacez=0isopen.
3 Supposev =x 3 i −3x 2 y 2 z 2 j + (7x +z)kandSisthe
surface ofacube ofside2units lying in the region
0x2,0y2,0z2withanopentopin
the planez = 2. Verify Stokes’theorem forthisfield.
4 Consideracube given by 0 x1, 0 y1,
1 z2,above the surfacez = 1. Suppose the
surfacez = 1 is the only open face. LetSbe the
surface ofthiscube. Verify Stokes’theorem forthe
field v =yi + (x −2xz)j −xyk. TakeC asthe square
withcorners (0,0,1), (1,0,1), (1,1,1), (0,1,1)in
the planez = 1.
5 Considerthat part ofthe positiveoctant, that iswhere
x,yandzare allpositive, bounded bythe planes
x=0,y=0,z=0andx+y+2z=2.Assumethat
the tetrahedron soformedhas three solidfaces, and
one open face onthe planex+y+2z = 2.Verify
Stokes’theorem forthe field v =x 2 i −2xzk.TakeC
asthe triangle (2,0,0), (0,2,0)and (0,0,1).
Solutions
When calculating the relevant line and surface integrals
the signofthe result dependsupon the orientation of
the curveC.
1 ∮ v·dr=± 1 2
2 ∮ v·dr=0
3 ∮ v·dr=±128
4 ±2
5 ± 2 3
27.9 MAXWELL’SEQUATIONSININTEGRALFORM
InSection26.7Maxwell’sequationswerestatedasexamplesoftheapplicationofvector
calculus.Infact,alternativeintegralformsoftheseequationsareoftenmoreusefuland
theseare given here forcompleteness.