3 3stokes 8 5場論定理Stokes設(shè)光滑曲面邊界是分段曲線

(Stokes定理StokesSL是分段光滑曲線,S的側(cè)L的正向符合右手法則,PQR在包含S在內(nèi)的一個(gè)空間域內(nèi)具有連續(xù)一階偏導(dǎo)數(shù),則有?R-?Qdydz+?P-?Rdzdx+?Q-?Pdxd

?z ?x ?yS n=GPdx+Qdy+Rd L

S:z=fx,y x,y?界曲線L在xOy的投影為有向曲線C。為此先證

znzn?Lox

dzdx-

dxd

=L1+f1+f2+fxy

,cosb

f

,cosg 1+1+f2+fxy1+f2+fxy

=-cosfx=-cosg

?Pdzdx-?Pdxdy

-?Pcosgd

S= ?P+?Pfcosgd =

y

fydxd

=-?Px,y,, =CP(x,y,z(x,y))dx=L

QdyL

=

?Qdxdy-

dydRdx S L ?R-?Qdydz+?P-?Rdzdx+?Q-?Pdxd

?z ?x ?yS =GPdx+Qdy+Rd=?R-?Qcosa+?P-?Rcosb+?Q-?Pcosbd

?z ?x ?y S dyddzddyddzddxd?PQRS =Pdx+Qdy+S ?Rdydz=?R ???PQR或用第一類曲面積分表示???PQR dS=Pdx+Qdy+ 例1.利 計(jì)算積分Gzdx+xdy+yd zdx+xdy+ydddyddzddxd???zxy=S

zΓo x=?y-?xdydz+?z-?ydzdx+?x-?zdxd

S zΓoDxy1x=dydz+dzΓoDxy1xS

+1dxd

z=1-x-D =111dxd =3dxdy=

11例2.G為柱面x2+y2=2y與平面y=z的交線 為順時(shí)針,計(jì)算I=G dx+xydy+xzdz2解:設(shè)為平面z=y上被G所圍橢圓域 cosb=1,cosg=- cosacosb xI=

d? ? ?d y2

xy 122cosacosb122

I=

? ?

d? d

,cosg=- y2

xy =

z-?xy

?y2

?xz

?xy

?y2cos ?y

+?z-

+ -?y d zcosys2=1 y-zdS=2z例3.利用 -z2x+2-x2y+2-y G 其中Gxyz

V=x,y,z)0￡x￡1,0￡y￡1,0￡z￡1}解:取平面x+y+z= 2

1,1, 3

2 3

2-z2x+2-x2y+2-y2z=

d S

-

z2-

x2-=

d

cosa=cosb=cosg=133132 33Sy2-z2

z2-

x2-

SS=6

2 cosa+ d?x2?z2 y2?x2 ? cosa+ d 3=-2+z+z++z+333=-4(x+y+zdS=-43d334343

43S43S S=-

23=-328.5 x,=x,ix,x,

S可知,單位時(shí)間通過曲面SSF=Pdydz+Qdzdx+RdxdSF=(Pcosa+Qcosb+RcosgdS v SF=Pdydz+Qdzdx+RdxdSSS當(dāng)F0時(shí)說明流入S的流體質(zhì)量少于流出的,表明S內(nèi)有泉;當(dāng)F=0時(shí),說明流入與流出S的流體質(zhì)量相等。 ,流量也可表為F=

+

+

dxdyd W ?zSM為了揭示場內(nèi)任意點(diǎn)M處的特性, 方向向外的任一閉曲面, 記S所圍域?yàn)閃, SM d點(diǎn)M(記作W d F=

1

+

+

dxdWfiM WfiM

W

?z+?,h,z+

,h,z+

R,h,z

1

dxdydWfiM

?y

=?P+?Q+?R

h,zW

M

Ax,=x,ix,x,

r, An0d,S

rdivA

rrAn0dS= 說明:由引例可知散度是通量對體積的變化率divA0表明該點(diǎn)處無源,r例如

vvvvvv為常數(shù)r rdivv=例1A(xyz)=(xyyexPxyQ=yex

求A(xyz)在點(diǎn)(01r|

?P+?Q+?R=y++ +=1+102divA(r例2.置于原點(diǎn),電量為q的點(diǎn)電荷產(chǎn)生的場強(qiáng)(rrx2+y2x2+y2+z

q E= r= x,y,

r

,r?0)?x+?y+?z divE=q?x 3

3 3 r2r2-r5

r2-3zr5+rr2-3zr5r53r2-3r= r5

= (

?0) r,

rotA=

-?z ?z

?x

-?y

?? ?y SrotAndS=At0d

Pdx+Qdy+Rdz=

At0d

wr=,0,w

r=x,y,z

M xv=·r

=wy,wx,rotv

????y-ww0????y-ww0例3.求向量場A(xyz)=(x2yzxy2zxyz2

ijk???ijk??? rotA=xz2-xy2,x2y-yz2,y2z-x2z

33,0i qxrEq=j?qyrrk?zqzrotE=rotE=例5

A=2y,3x,z2

:

+

+z2

4n為

計(jì)算ISrotAndS rotAr

??

=0,0,cosa\I=ScosgdS=2dxdy=x2+y2+x2+y2+ uurdivgradr=;rotdr= r- z ?x

r2-提示:gradr

?xr

= r ?y

r2-y2

?z

r2-z

r

(r3 (

r r3

divgrad

r2-x2

r2-+

r2- =+ =0rrrijkuur ??rot0rrrijkuur ??rotdr=xyzrrrA(xyzxyrdivAM=

A(xyzxy =r

F0,0mg, ?mg

divF=0+0 =

rotF=A(xyzxy

如果存在函數(shù)u(xy(xyz)∈V

A=?x,?y,?z r

rotA= A2yi2xj證