3 3stokes 8 5场论定理Stokes设光滑曲面边界是分段曲线

(Stokes定理StokesSL是分段光滑曲线,S的侧L的正向符合右手法则,PQR在包含S在内的一个空间域内具有连续一阶偏导数,则有¶R-¶Qdydz+¶P-¶Rdzdx+¶Q-¶Pdxd

¶z ¶x ¶yS n=GPdx+Qdy+Rd L

S:z=fx,y x,y˛界曲线L在xOy的投影为有向曲线C。为此先证

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或用第一类曲面积分表示¶¶¶PQR dS=Pdx+Qdy+ 例1.利 计算积分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为柱面x2+y2=2y与平面y=z的交线 为顺时针,计算I=G dx+xydy+xzdz2解:设为平面z=y上被G所围椭圆域 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,单位时间通过曲面SSF=Pdydz+Qdzdx+RdxdSF=(Pcosa+Qcosb+RcosgdS v SF=Pdydz+Qdzdx+RdxdSSS当F0时说明流入S的流体质量少于流出的,表明S内有泉;当F=0时,说明流入与流出S的流体质量相等。 ,流量也可表为F=

+

+

dxdyd W ¶zSM为了揭示场内任意点M处的特性, 方向向外的任一闭曲面, 记S所围域为W, SM d点M(记作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= 说明:由引例可知散度是通量对体积的变化率divA0表明该点处无源,r例如

vvvvvv为常数r rdivv=例1A(xyz)=(xyyexPxyQ=yex

求A(xyz)在点(01r|

¶P+¶Q+¶R=y++ +=1+102divA(r例2.置于原点,电量为q的点电荷产生的场强(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.求向量场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为

计算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

如果存在函数u(xy(xyz)∈V

A=¶x,¶y,¶z r

rotA= A2yi2xj证