微積分答案-第十章習(xí)題

10.3公式及其應(yīng)L2xyx2dxxy2dyLyx2y2x所圍成的區(qū)域的正LL2xyx2dxxy2L 12x3x2dxxx4dx202y3y4dy2y2y2 1QPdxdy

12xdxdy1

12xdy1xx

y

Lx2xy3dxy22xydy,L為由四個頂點(diǎn)分別為002022LLx2xy3dxy22xydyL 2x2dx2y24ydy0x28xdx0y QPdxdy

2y3xy2dxdy

2dx22y3xy2dy

y

xacos3t,yasin3 1

S2Lydxxdy2

a

tdacostacostda

t 8

9x216y2yL:yS1ydxxdy123sind4cos4cosd3sin122 2

x2y2x2acos2L:y2acosS1ydx2122acossind2acos22acos2d2acossin222

xacost,ybsint0t2S1ydxxdy12bsindacosacosdbsin2 2

r

y

cos;S1

ydx2 44

sinda

cos

cosd

sin

ydxxdyL為圓周x12y22LL 22L 2 1x2y2

x2 解

，所以取Cx2y22.則有

x2y2

2x2y2 ydxxdy

ydxxdy

2sindcoscosdsinL2x

2y2

C2x

2y2

2L2aydxxdyLxatsintya1costt02解：設(shè)直L12a000， L2aydxxdyLL2aydxxdyL2aydx

2dxdy2a2adx22a1costdatsint4a22 Lexsiny2ydxexcosy2dyL為上半圓周xa2y2a2y L解：設(shè)直L1002a0，Lexsiny2ydxexcosy2L1L1

exsiny2ydxexcosy2dy

exsiny2ydxexcosy

2dxdy2a0dxa2.0（1）

xydxxy

x2 解易得xydxxydyd xy所以曲線積分在整個xOy面內(nèi)與路 2,3xydxxydy5

3,46xy2y3dx6x2y3xy2 解：易得6xy2y3dx6x2y3xy2dyd3x2y2xy3，所以曲線積分在 面內(nèi)與路徑無關(guān)，3,46xy2y3dx6x2y3 （3）2,12xyy43dxx2 解：易得2xyy43dxx24xy3dydx2yxy43x，所以曲線積分在整個 xOy面內(nèi)與路徑無關(guān)，2,12xyy43dxx24 （1）L2xy4dx5y3x6dyL為三頂點(diǎn)分別為0030和32的L2xy4dx5y3x6dyD4dxdyLx2ycosx2xysinxy2exdxx2sinx2yexdy,其中LL x3y3a3a

x2ycosx2xysinxy2exdxx2sinx2yexdy

0dxdyL2xy3y2cosxdx12ysinx3x2y2dy,L為在拋物線2xy2上由L點(diǎn)00到,1的一段弧 L:,1,0;L:,00,1 2 L2xy3y2cosxdx12ysinx3x2y2LLL1

2xy3y2cosxdx12ysinx3x2y2

2xy3y2cosxdx12ysinx3x2y2 0

2 D0dxdy112y4ydy0dx4 2xL（4）x2ydxxsin2ydyL2xL1,1的一段弧；L1:1,11,0;L2:1,00,

上由點(diǎn)00到點(diǎn)xydxxsiny22LLL1x2xydxxsiny22LLL1x2ydxxsin2ydyx2ydxxsin2y 2xydxxydyL2

L D解 xydxxydy 2dxdy L Dxydyxydx其中L為圓周xya LL

a

解：LxydyxydxD

ydxdy

d0rdr 2xy2dxx2y2dy其中L為ABC的邊界，其中A1,1B32C3xy2dxx2y2

2x2ydxdy

33

x12x2y22 33x23x3

1（8）

4ex1cosydxysinydyL0x0ysinx解

x1cosydxysinydy exydxdy

exydy

1.

11x2xydxx2y2dyL為區(qū)域0x1與1y11解 x2xydxx2y2

xdxdy

0

1xdyexsinymydxexcosymdy,其中AMO為由Aa,0Ma,a2 2 至O00x2y2解：令L100a0，

exsinymydxexcosymdyexsinymydxexcosymdyAMOexsinymydxexcosymdy

0dx Fxy22xy8這變力確定了一個力場。證明質(zhì)點(diǎn)在此場內(nèi)移動時， 證明：易得x

dx2xy8dydxy 8y A計算曲線積 x44xy3dx6x2y25y4dyi1,2,3,其中A2,A1iA230,A30,3,A41,1AiAi1為任意的逐段光滑的曲線 2

2 5解：易得

dx6x

dyd2x

5

x44xy3dx6x2y25y4dy x44xy3dx6x2y25y4dy2913 x44xy3dx6x2y25y4dy2441 D是以逐段光滑曲線l為邊界的平面有界閉區(qū)域ux,yvx,yDucosn,xvcosn,yds

uvdl

y 其中cosn,x,cosn,y為曲線l的外法向量的方向余弦。此公式是 證明：設(shè)正方向切向量為，則cosn,xcosn,ycos,y,cos,x，于v,udsvdxudy vD y 曲線積分x22xydxx2y4dy是否與路徑無關(guān)？若與路徑其原函數(shù)L并計算由點(diǎn)O00B1,1LysinxL2

y5解：易得x2xydxxydydxy

.所以積分與路徑無關(guān)。 x22xydxx2y4dy23

52L:ysin 2設(shè)lrlcosr,nds其中nlcosr,ndsacosn,xbcosn,yl

r,ndsl

a2

dsD0dxdyI xyI xyn,其中l(wèi)n解：I x解：I xyn,

2dxdy2S.Sxdxydy2dxdy2S.Sx2xdxydyd1lnx2y2.x2 x0F

xiyj構(gòu)成力場，其中k為常數(shù)， x2x2 k 3xdxydyd fx在內(nèi)具有一階連續(xù)導(dǎo)數(shù)，L是上半平面y0內(nèi)的有向分段光滑曲線，其起點(diǎn)為ab，終點(diǎn)為cd。記I 11y2fxydxxy2fxy1Ly y2 IabcdI（1）Fxfx11y2fxydxxy2fxy1dydxFxyI y Ica

y2 PxydxQxydyxOy平面內(nèi)是某一函數(shù)uxy這樣的一個ux,y：（1）x2ydx2xyx2 解：x2ydx2xydyd 2xy 2xydxx2dydx24sinxsin3ycosxdx3cos3ycos4sinxsin3ycosxdx3cos3ycos2xdy2sin3ydsin2xcos2xdsin3ysin3ydcos2xcos2xdsin3ydcos2xsin3y3x2y8xy2dxx38x2y12yey3x2y8xy2dxx38x2y12yeydyydx34y2dx2x3dy4x2dy2d12yey12eydx3y4x2y212yey12ey2xcosyy2cosxdx2ysinxx2sin2xcosyy2cosxdx2ysinxx2sinydycosydx2y2dsinxsinxdy2x2dcosydx2cosyy2sinx.Xxy2,Y2xy8這變力確定了一個力場。證明證明x

dx2xy8dyd

8y（1）3x26xy2dx6x2y4y2dy

4y3解dx3ydx3xd

d 2

4y3

3dx3xy3 x33x2

4 3（2）a22xyy2dxxy2dy0(a為常數(shù)

y3解daxydxydxxdyxd

d y3

3daxxyxy3 3a2xx2y 33eydxxey2ydy解eydxxdeydy2dxeyy2xeyy2xcosycosxy'ysinxsinyxcosycosxdyysinxsinydxxdsinycosxdyydcosxsinydxdxsinyycosxxsinyycosxx2ydxxdy d3xy x3xy3

yx2ydxx2dyQ2xx4yP （7）1e2d2e2dde2dde2de20e2x2y2dxxydyQy2yP 確定常數(shù)，使在右半平面x0上的向量Ax,y2xyx4y2ix2x4y2為某二元函數(shù)ux,y的梯度，并求ux,yQ2x

4可解得u x2 由 x4y2,積分得ux,yarctanx2x.再2u

y2y1

'x

x4y2

得x0,xC.所以uxyarctan