2017高等数学辅导讲义练习题答案第九章解答

练习题精选答案与解析第九 【解】应选（D）.P(x,y)xay,Q(x,y) ,由题设知,PQ(x则(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则maxI1I2I3I4 .x2y2【解】应填【解】应填4πabc(a2b2c2(xyz)2dV(x2y2z2 z zdVczπab(1c2)dz15abc【解】应填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】应填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沿逆时针方向[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】应填2πa2a2a

2xz1)2dS(2x2z21

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

2x2z)dS

zdS D

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

【解】应填8.记为圆周(x1)2y21，则所求面积为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所围区域，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)从(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

其中为椭圆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.两边对t求导， 2t1 C(t)2t从而Cy2y1，所以Qxyx22y则f(xysiny

(t,t2而[sinyC(x)]dxxcosydyt tC(x)dxtsint2t0两边求导 C(t)2tsint22t2于是 f(x,y)siny2xsinx22x2cos【解】I dxxdy yf(xy)dxxf(xy)dL dxxdy dxcaL L F(x为f(xLyf(xy)dxxf(xy)dyLf(xy)d(xy)F(cd)F所以当abcdF(cdF(ab0Ica 30.（1）【证】 如右图所示，设C是右半平面x0内的任一分段光滑简单闭曲线，在C上任意取定两点M,N，作围绕原点的闭曲线，同时得到另一围绕原点的闭曲线.根据 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 设P φ( ,Q ,P,Q在单连通区域x0内具有一阶连续2x2

2x22x2y4总有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，将φy代入④得2y54cy32y5，所以c0，从而φyy231【解】1）由PQyg(xf(xf(xg(xex g(x)f(x)则

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 .【证】

ds cos(n,x)ds cos(n,L L ucos(τy)dsucos(τ （其中τ为曲线的切向量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其中为锥面z 的上侧，cosαx,cosβyx212I

2x

yDx]dSD

2x

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

0,

0,

0,xayaza

1a,(a13当xayaza a时，xyz 3a取最小值，最小值为3a,13(xyz3a)3dS(3a)3dS27a34πa2 【解】椭球面SP(xyz处的法向量是n2x,2yz,2z点P处的切平面与xOy面垂直的充要条件是nk (k22 yz即2zy0.所以点P22 yzx2 y2 D(x,y)

23y24

，记Σzz(xy),(xyDDIDD(xD

1x1xyzz y y2z2yz (x 3)|y2z4y2z24zz2

D3dxdy2πD

4y2z24y2z24dxdΣ1为xOy平面上被椭圆

1所围部分的下侧，记Ω为由ΣΣ14Ixzdydz2zydzdx3xydxdy(z2z0)dxdydΣ 03zd

dxdy6πz(1z)dzπ2x224

11I2xzdydz2zydzdx3xydxdy1

xydxdy2所 II1I2π

x1【解 设S为单位球面x2y2z21的内侧，由公式得xdydzydzdx

(x2y2z2)3/ 3(x2y2z2)23(x2y2z2)(x2y2z2)2dV(x2y2z2Vx y 其中为椭球面

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”号，当有向曲面的法向量指向内侧时，取“”号.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,

（其中是上半球体x2y22π

1r4sinφ

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