第七篇三重積分與線面積分僅數一考

§7.1f(xyz的連續(xù)性，則積分中值定理得I(rf(,,4r3其中3 f(0,0,0)，因此選r0 解：因為ez3tan(x2y3y的奇函數，且xoz[ez3tan(x2y3)]dv0，所以[ez3tan(x2y33]dv3dv3R2H 解：將

x2

1y2

by[ac(1 )]dyabcIydy0dyy

D(y解：將z軸，得投影區(qū)間[1,1]D(zx2y21z2

1ez[(1z2)]dz21(1z2)eI

dxdydz1dz

D(z02[ez(z22z2)ez]10y2x22zz4將z軸，得投影區(qū)間[0,4]D(zx2y22z0I4dz(x2y2z0D(z

2

4 (r zdz 解：將z軸，得投影區(qū)間[0,hD(zx2y2z20Ihzdzex2y20D(z f(x,y,z)

x2y2,0zx2yx2y2z2x2y

（2 0000 0000I d2

rr2 =51 解：積分區(qū)域被球面x2y2z21分成上內、外兩部分，分別記為 ， x2x2y2x2yx2y2

x2x2y2x2y2x2y22

d4

4 r4 )(

]sind(2 04cos43cos

( 4)(2 2) ( 解：F(t)[z2f(x2y2)]dv2d h[z2f(r 2tr[1h3f(r20

h3hf(t23 [limF(t)limF(t) 3 [t0t

ddf(r)r

4f(r)r

2f(r)r

0df(r)0

2f(r)

tf(r2) F(t)

2[t2f(t2)tf(r2)rdrtf(t2)tf(r2)r (tf(r) 02tf(t2)tf(r2)r(t 2 2 (tf(r)0所以在區(qū)間(0,F(t單調增加，故應選（C

zx2解：兩曲面的交線為xyz

x2xy2y

:(x1)2(y1)2 D故的體積為V1xyx2y2D3[3(x1)2(y1)2]dxdy [3u2v23

v2u22292d3r2rdr92192 解：(Ⅰ)由對稱性知的重心(x,y,z)(0,0,z)，按

1dz0z Dz D(zx2y20 1dz 所以z ，因此的重心(0,0,)

D(z(Ⅱ)拋物面x2y2z(0z1)上任意點(xyz2x(Xx)2y(Yy)(Zz)所以的重心與拋物面:x2y2z(0z1)上任意點(x,y,z)處切平面的距 2x22y2(2z)d

2x2x2x2y234x24y2tx2y2(t0)的重心與拋物面x2y2z(0z1)上任意點(x,y,z)處切平面的距離為d

2

(0t 2(2t)(14t) 2t 0,0td(t

0,t1

1 1

t 1所以t 6x2y2。 。z 離．在dv(xyzdv，則這一小塊可看成一個質量為(xy,z)dv在點(xyz又點(xyzxyz1,1,1x,y,3 (zy)2(xz)1,1,1x,y,3d ，所以d 3故xyzI(zy)2(xz)2(y

§7.2 (9x24y2xy2)ds(9x24y2)ds LLL24L

4)ds36L[4

x2y2z2⑵因為:xyz x2ds y2ds 所以x2ds1 3

(x2y2z2)ds又因為x2y2z2a2 (x2y2z2)ds ds 故

x2ds2a33I

ex2y2ds ex2y2dsex2y2ds ex2y2 y而在OAOAxx0xadsy，在上，xacost,0t ds，yasinx在OBOByx0x

42a，ds2

a

Iedx4eadt2e2x 2dx2(e1)ae x2y2z2

2I

(x2y2z2)ds

1 xyz (x)

2L: 因此L的參數方程可表為L:x 2cost,y2sint,z 2cos (0t 所以ds[x(t)]2y(t)]2[z(t)]2dtI(x2y2z2)ds9ds922ds18 2 2 uu4

．解：由于 ds= ，其中L Lxy

uu因此LdsL , xy u(L L LudyuL L(xy (xy 可得Lds[ 2 2]dxdy dxdy(1e)n

D y1

xy

D2dxdy4DCx12

xx12

dy

ydx(x1)dyx12

ydx(xx12 ydx(x1)dy

ydx(x1)dy 2dxdy x12y

D 2dy2sin22dysin2解：令C為過A(1,)、 2dy2sin22dysin2則ILCx2sin2

x2sin2

Cx2sin2

x2sin2 sinICx2sin2ydyx2sin2y dy sin dx dy sin ABx2sin2 x2sin2 BCx2sin2 x2sin2+ dy sin dx+ dy sin dxCDx2sin2 x2sin2 DAx2sin2 x2sin2P(xy)exsinyb(xy)Q(xy)excosy yb2I(ab)dxdy2abxdxa2(ba2 D

x軸圍成的區(qū)域，其面積為a22If(ycosxy]dxf(ysinx[f(y)cosxy]dx[f(y)sinx Dx根據題設dxdy2。又線段AB可用參數方程y2 D 1 1011 11 00I (xy)dx(xy)dy (xy)dx(xy)dy (xy)dx(x x2 L x2 x2=

(xy)dx(xy)dyx2y2 ，當t 時為開始，t 2(xy)dx(x 2I

41dt x2

4 Px2y2,Qx2y2 y2x ，(x,y)(0,0)⑴D:x2y2

x2yP

(xyD不是

0不是單連通區(qū)域，則

x2事實上，若取曲線Cx2y2r2 1ydxxdy 1 x

r2

ydxxdy 2dxdyr2r2

ydxx2 ⑵Dy0P

(x,y)D x2

ydxx2

(3xy2x3)dxP(xy)dyPI2

P(xy)dx3xy2x3)dy3x2(y)3y2因 P(x,y)3x2yy3c又因為P(0,1)1所以c0因此P(x,y)3x2yyC(y5ye2xf(x))dxe2xf(x)dyCx1e2xf(x2e2xf(x15e2xff(x3f(xe2xf(x1e2xe3xcf(0)6 f(x)1e2xe35I(2,3)y5ye2xf(x))dxe2xf2P(x,0)dx3Q(2,y)dy3e4(1e4e6)dy33e ，Q(x,y)3

3(x2y2) (x2y2) Q (x,y)(0, (x2y2)⑴ xdx恒有yx，所 3在域D:y0內是某二元函數u(x,y)的全微分(x2y2)Dx2y20不是單連域，令U表示全平面，則UDU0)．設C:x2y21，方向逆時針，則CDDxdxD

xdx3(x2y2)

xdxydy

(00)dxdy3(x2y2)

Dx

0內是某二元函數u(xyP(xy)2ax3y33y25Q(xy3x4y22bxy4 連續(xù)一階偏導數，所以它是某一函數全微分的充分必要條件為x 12x3y22by6ax3y26y，所以a2,b u(x,y)0P(x,0)dx0Q(x,

x5dxy(3x4y26xy

5xx4y33xy24y 2x2

l (xy

01 2x 2 2xP

(y)2x2y4

,Q

2x2y4

2 QP Q

2y(2x2y4)4x2xy(2x2y4

4x2y2 (2x2y4

(2x2y4

(2x2y4 (y)y4(y)y2y 由③得yy2c，將y代入④得2y54cy32y5所以c0，從而yy2

解：利用對稱性可得xds0將方程代入被積函數可得x2y2z2dsRds4R3 利用x,y,z的對稱性xds (xyz)ds ds x2y2z2a2關于三個坐標面都對稱，而函數2xy、2yz、2xz分別關于變量x、yz是奇函數，所以由對稱性可得 (2xy2yz2zx)ds2xyds2yzds 于 I(x2y2z2a2ds（因為在x2y2z2a2解：將積分曲面的方程表示成單值函數為：：z 積分曲面在xoy平面上的投影域為：D:x2y2R2，R2x2ds1(z)2(R2x2

2Rx2 Rr4sin2cos22R2r所以Ixyds= 2dxdyR0dR2r Rx R21cos

R5sin5drR2 R R2r R6242R6 解：令1為 Ixyzds4 11由于:zx2y2xoyDx2y21,x0,y01素 ds14x24y2所 I4xy(x2y2)14x24y2D 42

r5cossin14r211211

14r

drt

51151

( u2 udu (

1u3 32

5.解：由 n r5.解：由 n r I

ds

rndsr (xx0)dydz(yy0)dxdy(zz0 [(xx)2(yy)2(zz)2 當M0(x0,y0,z0)時，此時高 由 0，所以I0 當M0(x0,y0,z0)時，此時高斯 令：(xx)2(yy)2(zz)2 I11

(xx0)dydz(yy0)dxdy(zz030222[(xx0)(yy0)(zz)]02223(xx0)dydz(yy0)dxdy(zz0322 [(xx0)(yy0)(zz)]22020 [PQR]

(xx0)dydz(yy0)dxdy(zz00 022 [(xx022

(yy0

(zz)2]

(xx0)dydz(yy0)dxdy(zz03 [(xx)2(yy)2(zz)2 (xx0)dydz(yy0)dxdy(zz0上任意點(xyz

1 3dv x2n , x2

12x(x y(x 12所以I

12I [x(xy)y(xy)z2]ds (z2 x2y212 由于的方程為z ，它在xoy平面上的投影域為D:1x2y24，面元素為ds 因 I(x2y2D

x2y22d2(r2r)rdr21717 1解：令xex12(1x2)dydz8xydzdx

y2y2=

2(1x2)dydz8xydzdx4xzdxdy2(1x2)dydz8xydzdx1=(4x8x4x)dxdydz2(1x2)dydz8xydzdx =[2(1x2)dydz8xydzdx+4xzdxdy]=2(1x2 =2(1e2)dydz1

y2z2

2(1e2)dydz2(1e2)解：記圍成的區(qū)域為 ， ，Q ，R

3(x2y2z2) (x2y2z2) y2z2 x2z2

(x2y2z2) y2x22zx 5，(x2y2z2)

5，(x2y2z2)

5(x2y2z2)當(0,0,0)時，P,Q,R在內有連續(xù)一階偏導數， I(PQR)dv 當(0,0,0)P,QR在內除(0,0,0)1:x2y2z221 I xdydzydxdzzdxdy xdydzydxdz (x2y2z2)1

333dvx2 在所圍的立體內有偏導數不存在的點（zx2x2y令1為在錐面zx2y1D:1x2y21令為z2xoyDx2y2 令為z1xoyDx2y2 z則I z

dxdy

zz zz

ee e

2

2e2

1 rdr rdr

1 0 013因此I [fx2fyfz]ds (xy13

ds1313xx3 rotF

6xyy,13y,3x 所以rotFds(6xyy)dydz1

)dxdz3xx2y令1為平面z0在錐面zx2yrotF

(6xyy)dydz(13y2)dxdz3x (6xyy)dydz(13y2)dxdz3x1(6y6y)dV3x2dxdy3x 3x2dxdy (x2y2)dxdy32r3dr22 x2y2

xy

na2,b2,c2 所以

(z)0，即 z 22 22 22I

xdydzydxdzzdxdy

c (xy

由于