2017高等數(shù)學輔導講義練習題答案第九章解答

練習題精選答案與解析第九 【解】應選（D）.P(x,y)xay,Q(x,y) ,由題設知,PQ(x則(a2)xay2y,(a2)xa2)y故a

(x 公式知Ii(1

2)dxdyDiLi (i1,2,3,4DD上，1x2

y22

0,

D1,0I1

I2I4I3I4則maxI1I2I3I4 .x2y2【解】應填【解】應填4πabc(a2b2c2(xyz)2dV(x2y2z2 z zdVczπab(1c2)dz15abc【解】應填π(1 I 2πdθ1dr1re(1z)2dz1dz2πdθ1ze(1z)2dr (a2b21)L原式(b2x2a2y2C22a2b2(22C

y

)ds(a2b24πa33[(x2)2(y3)2]ds[(x2y2)(4x6y) (xy)ds

3(xyz)ds

ads (4x6y)ds2yds2(xyz)ds20ds 3 313ds【解】應填0LL1L2L1y1x(1x0)L2y1x(0x1xydxx2dy0[x(1x)x2]dx0(2x2x)dx1， xydxx2dy

2 [x(1xx]dx(x2xdx6，故Lxydxxdy ILx(xy)dxxdy

y(dxy2

（L

1沿逆時針方向[x(xy)y2]dx[xy2 (1x x2y2

.

2 ds

cos(n,x)ds

ucos(n,L L [ucos(τ,y)dsucos(τ,L D DL[

dyx

ydx]

x2

y2 D

(

y

)dσ2

（Dx2y218πR4原式x2z2dS2x2y2z2 38πR3【解】應填2πa2a2a

2xz1)2dS(2x2z21

2xz22x(2x2z2)dS1(2x2z2 （其中1x2y2z2a2211(x2y2z2)dS1a2dS 1dS (22xz

2x2z)dS

zdS D

1z2z2D2adxdyD【解】應填(2 2)πR3. 公式4xdydzydzdxzdxdy3dv32πdθπdφRr2sinφdr(24

【解】應填8.記為圓周(x1)2y21，則所求面積為S

x2y2ds0

4cos2

θ4sin

θdθ82cosθdθ11

zz

dV，而

dV

dxdyz

πzdzπ2

d dx2y2

且zdV dxdyzdzπzdz，故z 0 x2y2 0

【解】x(1z)dvxdv1xπ(xx2)dxπ 【解】|zx2y2|dv(zx2y2)dv(x2y2 其中zx2y2z1所圍區(qū)域，其中zx2y2x2y21z0 11(zx2y2)dv (zρ2)ρdz11 2π6 6 ρ (x

z)dv

dθdρ

(ρz)ρdz

【解】1dx1xdy1xy(1y)e(1yz)2dz1dy1ydz1yz(1y)e(1 1dy1y(1y)(1yz)e(1yz)2dz11(1y)[1e(1y)2]dy1 0 .I 3x2ydx(x3x2L 3x2ydx(x3x2y)dy 3x2ydx(x3x2 L (x3x2y)(3x2y)]dxdy

( D D D

220

42

2BAIL π(siny3siny)dσ(siny3siny)dσ

(ex2sinx ππ dxπ a2a2

,L

(a0)從(a,0)到(a,0) (xy)dx(x x2

(xy)dx(xy)dyx2y2

1a21

(xy)dx(x1a2L1

(xy)dx(xy)dy

(xy)dx(x 2dxdy2

xdxπD a2D【解 由于P (x2y20)， xdyydx

xdyL

4x2

其中為橢圓4x2y214x xdy4x

xdy4x2

xdy4x2 xdyydx 21dy

7 04 【解

(2xy)Q(xy)x2Cy，其中Cy(t2xyd Q(x,y)d

12C(y)]dyt21C(y)d Q(x,y)d

2C(y)]dyttC(y)d t21C(y)dyttC(y)dy.兩邊對t求導， 2t1 C(t)2t從而Cy2y1，所以Qxyx22y則f(xysiny

(t,t2而[sinyC(x)]dxxcosydyt tC(x)dxtsint2t0兩邊求導 C(t)2tsint22t2于是 f(x,y)siny2xsinx22x2cos【解】I dxxdy yf(xy)dxxf(xy)dL dxxdy dxcaL L F(x為f(xLyf(xy)dxxf(xy)dyLf(xy)d(xy)F(cd)F所以當abcdF(cdF(ab0Ica 30.（1）【證】 如右圖所示，設C是右半平面x0內的任一分段光滑簡單閉曲線，在C上任意取定兩點M,N，作圍繞原點的閉曲線，同時得到另一圍繞原點的閉曲線.根據(jù) 2x2φ(y)d 2x2

φ(y)dx2xyd2x2

φ(y)dx2xyd2x2y

φ(y)dx2xyd2x2

φ(y)dx2xyd 2x2 φ(y)dx2xydy φ(y)dx2xyd 2x2 2x2

φ(y)dx2xyd2x2y

φ(y)dx2xyd2x2

2 設P φ( ,Q ,P,Q在單連通區(qū)域x0內具有一階連續(xù)2x2

2x22x2y4總有QP Q2y(2x2y4)4x2xy4x2y2 (2x2y4 (2x2y4)2 φ(y)(2x2y4)4φ(y) 2x2φ(y)φ(y)y44φ(y) (2x2y4 (2x2y4

φ(y)y44φ(y)y32 ④由③得φyy2c，將φy代入④得2y54cy32y5，所以c0，從而φyy231【解】1）由PQyg(xf(xf(xg(xex g(x)f(x)則

g(x)g(x)exf(x)g(x)g(x)CexCex1 f(x)g(x)(C1)exCex1 由f(0)g(0)0C1C1 f(x)1(exex)1 g(x)1(exex)1 I21[yg(1)f(1)]dy1(7e .【證】

ds cos(n,x)ds cos(n,L L ucos(τy)dsucos(τ （其中τ為曲線的切向量L

dyx

ydx]

Lx2y21I(x2y2)(dyfdx)f

dyf L 2 2D( y2D I(x2y2)(dyf

22 2f2(xxyy)dσ(xy)x2y2 DD222222 (x2y2)dσ2(xxyy)dσ(xy)x2y2D

D

D2

2f (xxyy)dσ2[1(xy)]x2y2 1[1(x2y2)]e(x2y2)dσ12πdθ1(1ρ2)eρ2ρdρ2coscoscosyzxI

ydxzdyxdz 1313

dS(3)

πa2

1313IydxzdyxdzCydx(xy)dyx(dx((yx)dx(2xy)dy(21)dxdy(3)

13πa213CD其中Cx2y2z2a2xyz0xoyIxydxz2dyzxdz

cosz2

[2zcosαzcosβxcosγx212其中為錐面z 的上側，cosαx,cosβyx212I

2x

yDx]dSD

2x

yDx]2dσxdσD2 2 2 222223【證】令Fxyz 3aλ(x2y2z22ax2ay2az2a23

0,

0,

0,xayaza

1a,(a13當xayaza a時，xyz 3a取最小值，最小值為3a,13(xyz3a)3dS(3a)3dS27a34πa2 【解】橢球面SP(xyz處的法向量是n2x,2yz,2z點P處的切平面與xOy面垂直的充要條件是nk (k22 yz即2zy0.所以點P22 yzx2 y2 D(x,y)

23y24

，記Σzz(xy),(xyDDIDD(xD

1x1xyzz y y2z2yz (x 3)|y2z4y2z24zz2

D3dxdy2πD

4y2z24y2z24dxdΣ1為xOy平面上被橢圓

1所圍部分的下側，記Ω為由ΣΣ14Ixzdydz2zydzdx3xydxdy(z2z0)dxdydΣ 03zd

dxdy6πz(1z)dzπ2x224

11I2xzdydz2zydzdx3xydxdy1

xydxdy2所 II1I2π

x1【解 設S為單位球面x2y2z21的內側，由公式得xdydzydzdx

(x2y2z2)3/ 3(x2y2z2)23(x2y2z2)(x2y2z2)2dV(x2y2z2Vx y 其中為橢球面

1x2y2z21 (xyz 23/ xdyd(xyz 23/

xdydzydzdx(x2y2z2)3/ SxdydzydzdxS (111)dvx2y2z20[yf(x)yz]dydz1y2f(x)dzdx[zysinxy2]dxd V(yf(x)yf(x)ysinx)dV其中VSS”號，當有向曲面的法向量指向內側時，取“”號.由S的任意性，知f(x)f(x)sinxf(0)0,f(01f(x)3ex3ex1sin （2）由（1）可知yf(xyz]dydz1y2f(x)dzdxzysinxy2 [yf(x)yz]dydz1y2f(x)dzdx[zysinxy222其中2z1zx2y2[yf(x)yz]dydz1y2f(x)dzdx[zysinxy2]dxdy [ysinxy2

(x2y2)dσπ

x2y22x2y2 x2y2【解 取曲面1:z

x3dydzy3dzdx(zf(x,y)z3x3dydzy3dzdx(zf(x,y)z3x3dydzy3dzdx(zf(x,y)z3[3(x2y2z2)f(x,

（其中是上半球體x2y22π

1r4sinφ

f(x,y)dv2 2f(x,y)dv f(x,y)2xy2x2