微積分3sadfa第四次習題課題解

30S1．S為球面(xa)2yb)2zc)21，則S

(xyz)dS(4(abc)S(xyz)dSS[(xaybzc)]dSS(abc)dS2．S為平面x2zSdzdxdxdy(

在柱面(x10)2y10)21內部的部分上側，則SdzdxdxdySdzdxSdxdySdx (x10)2(

dxdySSxyz4x2y21S

zdS(

dS

3dxdy．SzdS

x2y2

(4x

3dxdy

3LLx2y29，順時針方向.，則L

(2xy2y)dx(x24x)dy(18提示：S為0x,yz1

x5dydzy6dzdxz7dxdy2Lyx21x1，則|x|dl（1(5312 S:x

yz1(xz)dSSyozxdS0S

zdS1(x

yz)dS1dS36 3

3 (xz)dS36 x2y２．設x2y2azzx2yf(xyz)dV

（a0Ia

a2a2da2a2

f(x,y,

x2aI2dardr2arf(rcos,rsin,0I2d

ra f(sincos,sinsin,cos)

204

sin2

f(sincos,sinsin,cos)

xyz)dy^dzxy)dx^dySx2y21(0z2，外側vn解法１：x(y (xy)k，xiyj，dSddz（02,0z2vnSx(yz)dy^dz(xy)dx^

SvndS

(yz)ddz

cosd0(sin-z)dz解法２：:x2y21,0z

2

Sv

S1v

S2vS1x2y21z0S1x2y21z2

0,

0

S2

(xy)dy(x計算 x2y （１）Lx2)24y3)21 （２）Lx3y31（1）Lx2)24y3)21Y

y2x2(x2y2

，由，(xy)dy(x x2y

（2）L:x3y31，順時針方向，包含原點，不能直接用1Lx2y221(xy)dy(xy)dx=

)

=

2 x2y

a2b2c2f(axbycz)dSa2b2c2

u)duSS:x2y2z2

f(x,yza2b2c證：設uaxbyczvk1xk2yk3zwm1xa2b2c換．在uvwS的方程為u2v2w21,vsincos,wsinsin,ucos．02,0a2b2c2f(a2b2c2u)dS2a2b2c2

cos)sinS0 0

fx,yC(Dgx,yR(Dgx,y)DR2為有界閉域，則存在(,Df(x,y)g(x,y)dD

f(,)g(x,D證：fC(D),mfxyMxyD.設g0，mg(x,y)f(x,y)g(x,y)Mg(x,y)，gR(D)mg(x,y)df(x,y)g(x,y)dMg(x, fm cM(,Df(,)c，g(x,DfgdD

f(,)DfxC[ab],x[ab],fx)0bf(x)dxbdxba)2 af(x) b

bIaf

f(y)af(

f(x)2If(x)df(y)d(f(x)f(Df(f(x)

Df(

f(1

f(

f(D

f(

]2

f(

]2d2(ba)2Ibfx在區(qū)間[0,1上連續(xù)，證1dx1dyyfxfyf(z)dz1(1fx證設0f(t)dtFx

0dxxdyxf(x)f(y)f(z)dz0f(x)dxxf(y)F(z)1f(x1xf(1f(x1xf((y)1f(1F2f(1F0221F)12(0221

xd (y)F(x)]xdx1111

F2(x)F(1)F)F(1)F(x)]dF([2

F2(1)F(x)

12

F3(x)

1 11F3(1)1(1f fxf(xy)dD

(2auf(u)duD:xa,ya已知函數(shù)f連續(xù),試 xdxxdxf(x)dx x(x2y2)nf(

(2n)!! n

x

f(x)dxxzdzxf( 交換次序

x 00 00

f(y)dyyzdz

(x

)f(nn+1左邊xxdx x(x2y2)n1f(y)dy (2(n1))!!x交換次x

f(y)dyy

x12（nx121xx2y2nfy)dy2n!!2(x2y2h(ttzh(t

0.9，問高度為130（厘米）解:(

zh(t)

2(x2y2

D{(x,y)

12

4xh(t),y1(z)21(z)2(z

h2h2(t)16x216h2 1

d

1h(t)

h2(t)16r2h(t) 1

21 h(t)

(t)

22 22

Vd