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

27.8 The divergence theorem and Stokes’ theorem 895
EXERCISES27.7
1 Given F = −yi −xk evaluate ∮ SF·dSwhereSis the
surface ofaunitcube 0 x1, 0 y1,
0z1.
2 Given F = 6xy 2 z 2 k evaluate ∫ SF ·dS whereSisthe
plane surfacez = 2, 0 x2, 0 y2. Take the
direction ofthe vector element ofarea to bek.
3 Evaluate the volume integral ∫ V 6xdV whereV isthe
parallelepiped 0 z2,0 x1, 0 y3.
4 Evaluate ∫ V 1 +zdV whereV isaunitcube
0x1,0y1,0z1.
Solutions
1 0
3 18
2 128
4
3
2
27.8 THEDIVERGENCETHEOREMANDSTOKES’THEOREM
Thereareanumberoftheoremsinvectorcalculuswhichallowline,surfaceandvolume
integrals to be expressed in alternative forms. One of these, Green’s theorem, has been
describedinSection27.6.3.Thisallowsalineintegraltobewrittenintermsofadouble
integral.Now wegive details ofthe divergence theoremand Stokes’ theorem.
27.8.1 Thedivergencetheorem
The divergence theorem relates the integral over a volume,V, to an integral over the
closedsurface,S, whichsurroundsthatvolume, asillustratedinFigure 27.23.
When calculating such surface integrals vectors drawn normal to the surface should
always be drawn in an outward sense, that is away from the enclosed volume. Recall
thatwhen a surfaceisclosedthe symbolforasurfaceintegralis ∮ S .
Thedivergence theorem:
∮ ∫
v·dS= divvdV
S V
Example27.18 Verify the divergence theorem for the vector field v = x 2 i + 1 2 y2 j + 1 2 z2 k over the unit
cube0x1,0y1,0z1.
Solution Firstlyweneedtoevaluate ∮ S v·dSwhereSisthesurfaceofthecube.Thisintegralhas
been calculated inExample 27.17 and shown tobe2.
Secondlyweneedtocalculate ∫ divvdV overthevolumeofthecube.Thishasbeen
V
done inExample 27.16 and again the resultis2.
We have verified the divergence theorem that ∮ S v·dS=∫ V divvdV.