同濟(jì)高等數(shù)學(xué)課后習(xí)題解答下冊(cè)

第八 第一 多元函數(shù)的基本概

(x,y)(x0,y0

f(x,y)A是點(diǎn)(x,y以任何方式趨于(x0,y0習(xí) f(x,y)xyyxf(xyx解f(xyxyxyxyxf(xyxy)x2y2f(xf(xyxy)xy)(xyf(xy)x1x1x2yzln(xy1)xy11

2

xy0x2y20x zln(x22yx22y1f(x,y)ln(1|x||y解：1|x||y|0|x||y|

(x,y

1xxyx2y2解：(x,

1xxyx2

2 xy(x,2 xy2 xy2 xy1xy4

8

(x,y 22 xy

(x,y 44(xyxy(2 xy

(x,y)(0,0) (2 xy(2 xy

(x,y (x,y (x,y

(x,

(2

(4)

x2y21x2y21

(2x)sin(xy)

lim[(2x)sin(xy)x]

(x,y (2

(x,y (2x)xy

(2x)x(x,y x2y21

(x,

(x,x(4)x x2yx2y21

y

1

x2

1lim(x2

)

x2x2y21x2y21

2x0x2 (x2 (x2y2)(x2y21

2222x2

x2 x2y2 x2y212

x2 證明下列函數(shù)當(dāng)(xy)0,0x2yf(x,y)

x2y

x2

x2k2

1kxx0xf(x,y)

x0x2k2x2

1k

x2y2(xx2 yx0x2y2(xy

x0

x20x0x2y2(x

z x解xy2(2)z

y2y2第二節(jié)偏導(dǎo)zf(xy在(x0,y0fxfy

,,

))

f(x0x,y0)f(x0,y0)f(x0,y0y)f(x0,y0)f(x,

的幾何意義為曲線zf(xy在點(diǎn)M(x,

,f(x,

x

y

0 0f(x,y)在任意點(diǎn)(x,y處的偏導(dǎo)數(shù)fx(x,y)、fy(x,yfx(x,y時(shí)，只需把yxzf(xy的偏導(dǎo)數(shù)fx(xy),fy(xy42z2 2 2x2y2xyyxzxy

習(xí)題z1yzx zarctanx

(x(

(1y (

x2y2

1y

x2zln(x

x2y2 x x2x2x x2x2x2z

x x2x x2y(x x2y2)x2uln(x2y2z2u

,u

2 ,u x2y2u yzet2

x2y2

x2y2x

x2

,u

y2

,u

y2

x2zsinxcos

解 x

z(1xy)xz1xy)xy[ln(1xyx

y],u(1xy)xy[ln(1xy)xy

1

1uecos(sin(uecos(sin( xyxyzx2(y1)

z

x解：z(0,1) xx0zx2eyx1arctany

yz(10y

ey

2zxln(xy) xzln(xy1z

2y

2，2z

2

2

x y2 2

2

2

2zcos2x2y，

x

y

x2cos(x2ysin(x2y)sin2(x2z4cos(x2y)sin(x2y)2sin2(x2 2

2

8cos2(x2

2

4cos2(x2z

x2y2

2etdt， x

2z

x2

ex

2

2(1

)ex2

ex

2

x2x3y

x2y2f(xy)

x20

x2y2

fxy(0,0fyxfx(00

x0f(0,0)

f(0,y)f(0,0)lim00 y0x44x2y2 fx(x,y)

(x2y2

,xyx44x2y2 fy(x,y)

(x2y2

,x

y5f(0,y)

f(0,0)lim lim f

(x,0)

f(0,0)lim lim (11

2

2z

xy

x

y解:z

(11)exy

(11ex

(11

(11

(11x2 y2 x2

ex

y2

ex

2exy

x2y2x2y2z

x

y

z2

rx r

r2

x

x2y2

r r2

r2

2rr2

r2, 2r2r2r2r2x2y2z2r2

第三 全微z

f(x,y)在點(diǎn)(x0,y0處的全增量zx2yzAxByx2y則稱(chēng)z

f(xy)在點(diǎn)(x0,y0)可微，并稱(chēng)AxByAdxBdy為z

f(xy)在點(diǎn)(x0,y0的全微分，記作dzzf(xy在(x0,y0（1）f(xy在(x0y0

（2）f(xy在(x0,y0A

fx(x0,y0),B

fy(x0,y0dz

fx(x0,y0)dxfy(x0,y0)dy

dz

fx(x,y)dxfy(x,y)dyzf(xy在(x0,y0的某鄰域內(nèi)可偏導(dǎo)，且偏導(dǎo)數(shù)在(x0,y0zf(xy在(x0,y0習(xí) x2x2y

zarctanx1解:dz

dln

2

1d(x2y2)

xdx zarctanx1

x2

x2解:dz

1(xy1

dx1 (1 (1xy)(dxdy)(xy)(ydxxdy)(1y2)dx(x2(1xy)2(x

(1

(1xy)2(xzysinx y解:dzdesinxlnyesinxlnyd(sinxlnyysinxcosxlnydxsinxyx2x2yx2解:dux2

x2

x2y2dzx2y2dzzdx2x2y2dzzxdxx2(x2y2)dzz(xdx3(x2y2uex(x2y2z2解:dudexx2y2z2)exx2y2z2)d[x(x2y2z2d[x(x2y2z2)](x2y2z2)dxx(2xdx2ydy(3x2y2z2)dx2xydy所以dudex(x2y2z2)ex(x2y2z2)[3x2y2z2dx2xydyu解:dudxyzdeyzlnxeyzlnx(yzdxzlnxdyylnxxyz(yzdxzlnxdyylnxdz)zln(1x2y2x1y2dz

dz2(xdx1x22(dx2dy)2(dx

11 zyxxdy

x2y1x0.1y0.220.2解:dz

dz|(2,1)

4zy

y

0.811.62.10.5x(201,10

x

(x,y)f(xy)

x2

(x,y)

在點(diǎn)(0,0)解:由于limf(x,y)lim(x2y2)

0f(00f(xy在點(diǎn)(00

x2x2sin1f(00

f(x,0)f(0,0)

limxsin1

y2

1

yf(0,0)y

f(0,y)f(0,

limysin12 2f(xyf(00x2y2

x2x2x2f(xyf(00fx(00)xfyx2

x2

f(x,y)f(0,0)fx(0,0)xfy(0,0)y x2x2x2

x2f(xy在點(diǎn)(00(1.02)3(1.02)3f(xy)

x3，則dfx3

3x2dx x3再設(shè)(x0y012x0.02y(1.02)3 f(xx,yy)f(x,y(1.02)3 13 30.0212(0.03)30.060.3613213x2x6m,y8m解：對(duì)角線長(zhǎng)為fx2

xdxx2xdxx2f(6.057.9f(6,8df|(6,8)

60.0580.1100.5626262第四節(jié)多元復(fù)合函數(shù)的求導(dǎo)設(shè)u(x, v(x,y)在(x,y)可偏導(dǎo)，z

f(uv)在相應(yīng)點(diǎn)有連續(xù)偏導(dǎo)數(shù)，則zf(xy),(x,y)在(x

zfu

v

zfuf u

v

u

v多個(gè)中間變量：設(shè)u(x, v(x,y)

w(x,

zf(uvwzf(x,y),(x,y),(x,y)zfu

vfw

zfu

vf u v w

u v w只有一個(gè)中間變量：設(shè)u(x

zf(xyuzfxy,(xy)zfuf zfu u u 只有一個(gè)自變量：設(shè)u(tv(tw(tzf(t),(t),(t)dzfdu

dvf u

v w

習(xí)題zex2y

xsin

ydzzdxzdyex2ycost2ex2y3t2cost6t2 x yzarcsin(x

x

y31(x131(x1(x312t1t2(34t2解 x y

z y z

(x解 y

1 1 1u

eax(y,a2

yasin

zcosduuudyudzaeax(yz)eaxacosxeaxsin

y

z

1

1

11

sinxacosxacosxsinx)

1

1)sinx

sinzu2 ux

vxx2u2v2(uvz2u2v2(uv)4zx2ln

xst

y3s 2解：z2x1lny3 22

s[2ln(3s2t)

t2

t2(3s t

3sz

lny

[ln(3s2t) t2

t2(3s t

3szf(x2y2,exyz2xfyexyfz2yfxexyfz

f(xy,zyfzxffz

yf(x2y2z

2xyf

f2y2f， fuxyzf(yx

fuyzfyyyzfzxzf1xzfu 設(shè)uf(xxyxyzfxufyfxzf

xf12zxf13fy[xfzxf]zfyz[xf

zxfx2yy)f,有二階連續(xù)導(dǎo)數(shù)，求xzfxfy2y1fyf 2z f f

f2x 2z

f(xy,ygxfg有連續(xù)二階偏導(dǎo)數(shù)，求zyf1fgyyf

fy y

y 2

fxyfxf

fxf

f

g x1x1

y

y2 y

y3

x311

f2

gyg第五節(jié)隱函數(shù)的求導(dǎo)公F(xy)0yy(x，則dyFx F(xyz)0zz(xyzFxzFyG(x,y,z)（1）若F(x,yz)0確定yG(x,y,z)

dy

(F,(x,(F,(y,

，dz

(F,(y,(F,G)(y,（2）若F(x,yuv)0確定uu(x,y)G(x,y,u,v) vv(x,(F, (F,

(F,

(F,(F,(F,(F,(F,(F,(u,(u,(u,(u,

(x,

，u

(y,

；v

(u,

，v

(u,y)習(xí)題8—yy(xx2xyey2xdxydxxdyeydy0eyx)dy2xy)dxdy2x eysinyexxy2解sinydyexdxy2dx2xydy0siny2xy)dyy2exdy

y2sinyxyyylnxxlnylnxdyydxlnydxxdyxylnxdyy2dxxylnydx x(ylnxx)dyy(xlnyy)dxdyy(xlny x(ylnx

arctanx2x2yx2yarctanyxdxx2y

xdyydxxdy x2xdxydyxdyydxdyx x

1( x

x2zz(xy的一階偏導(dǎo)數(shù)zxz32xzy解z32xzy03z2dz2zdx2xdzdy03z22x)dz2zdxz

3z22x

,z

3z22x3sin(x2yz)x2y3sin(x2yz)x2yz3cos(x2yz)(dx2dydzdx2dy[3cos(x2yz)1]dz[13cos(x2yz)](dxz1,zxz

lnyxzlnzzlnydx1lnz)dzlnydzzyy(1lnzlny)dzydxzdy，z

,z x2yz

1lnzln1

y(1lnzlnx2yz

0dx2dydz (yzdxxzdyxydz)z

xy)dz(yzyz xyzyz xyzxz xyz

xyz)dx(xz

設(shè)ezxyz

解ezxyz0ezdzyzdxxzdyxydz0ezxy)dzyzdxz ,z ezxy ezzz(ezz (1zez)z

(1zez) zy2z

y

(ez

(ez

ez(ez

xxzezyez(ez

z(1xyz2y2zy2)x2y2(z1)3

z33xyz

求解z33xyza33z2dz3yzdxxzdyxydz0z2xy)dzyzdxz ,z z2xy z2(zyz)(z2xy)yz(2zz (z

)(z2xy) 2z

(z2

z2 z2(z2[z(z2xy)yxz](z2xy)yz[(2zxzx(z2xy)]z52xyz3x2y2z(z2 (z2設(shè)exysin(xz)1

求解exysin(xz1exysin(xz)(dxdyexycos(xz)(dxdzcos(xz)dz[sin(xz)cos(xz)]dxsin(xztan(xz)1,ztan(x2z

sin(x

sec(x sec(xz)tan(xz) cos3(x設(shè)zlnz xet2dty

求zlnzxet2dt011)dzex2dxey2dy zex2 zex1z,y1

ze (1

x2ze

ye

1

(1

(1

(1設(shè)uxy2z3zz(xyx2y2z23xyz所確定的隱函數(shù)，求

解uxy2z3duy2z3dx2xyz3dy3xy2x2y2z23xyz2xdx2ydy2zdz3yzdxxzdydu|(1,1,1)dx2dy3dz|(1,1,1)2dxdyx(1,1,1)zx2y（1）設(shè)x22y23z220

dy,dz

dz

dz2xdx2

dz2ydy

1 dy2xdx4ydy6zdz03zdz2ydy dy

x(16z)dz

x1

,dy

x(16z)2y(13z)

2y(1xeuusin (2)設(shè)yeuucosv，求x,y,x,dyeucosv)duusindu

usinvdxucosvdy dv u[eu(sinvcosv)u

,u

cos eu(sinvcosv)

eu(sinvcosv)

cos

,v

eu(sinvcosv)

eu(sinvcosv)xeucosv,yeusinv,zuv，求zx dyeusinvdueucosvdvdveu(sinvdxcos 又dzvduudvveu(cosvdxsinvdyueu(sinvdxcoseu(vcosvusinv)dxeu(ucosvvsinzeu(vcosvusinvzeu(ucosvvsin yf(xt，而tF(xyt)0xyfF

fFfdyx t fFy

t f(x,t)，dyf1dx F(xyt0FdxFdyFdt0dt

1(FdxFF F 3dy

fdxfFdxfFdy(FfF)dy(FffF2 222 22

2 3 2 FffF所 3 2 2 Ff 2

第六節(jié)多元函數(shù)微分學(xué)的幾何應(yīng) 設(shè)點(diǎn)M0(x0,y0,z0) 若:xx(t),yy(t), 則切向量為

x(ty(tz(t；其中x2(ty2(tz2(t0

xx0x(t0

yy0y(t0

zz0z(t0

x(t0)(xx0)y(t0)(yy0)z(t0)(zz0)0

:F(x,y,z)0G(x,y,z)則切向量為τ(F,G)(F,G)(F,G)

MM(x,y,z

(0)(y,

(z,

(x,y)00

FxFyGx

00GzM(x,y,00

x

y

z (F,(F,(y,M(F,(z,M(F,(x,M

(xx0)(F,(F,(y,

(yy0)(F,(F,(z,

(z(F,(F,(x,

)0 設(shè)點(diǎn)M0(x0,y0,z0) 若:F(x,y,z)0則法向量為nFx(M0Fy(M0Fz(M0F(M0)(0

Fx(M0)(xx0)Fy(M0)(yy0)Fz(M0)(zz0)0

x Fx(M0

yy0Fy(M0

zz0Fz(M0 若:zf(x,y)則法向量為nzx(x0,y0zy(x0,y0),1

zz0zx(x0,y0)(xx0)zy(x0,y0)(yy0)

x zx(x0,y0

yzy(x0,y0

zz0 若:xx(u,v),yy(u,v),zz(u,v)

(y,z)(z,x)(x,y)則法向量n

(u,v),(u,v),(u,v)

(0)0z 0z

v(u0,v0

(u,v

(y,((y,(u,

(xx0)

(z,(u(z,(u,

(yy0)

(x,(u(x,(u,

(z

)0

x

y

z

0(y,(y,(u,(u0,v0(z,(u,(u0,v0(x,(u,(u0,v0x1tytz

對(duì)應(yīng)t1的點(diǎn)處的切線和法平面方程 1 解： ,2t) ,t2(1 yx切線

2z 法平面：4(x2)y 8(z1)04xy8z （1）ezzxy3，點(diǎn)解：n(y,x,ez1) x22y10x2y

x2y1 x2 y（2） a

，點(diǎn)(x0,y0z0 2x2 解： )

(2x0,2y0,a2 c(x0,y0,z0

2x0(xx2y0yy1(zz

2

2 0(xx) 0(yy) 0 0

2xx0(xx2yy0yyz

x y z a2(xx b2(yy c(zz法線： 0 0 0 0 0

c

2 xt3yt2ztx2yz6txt3,yt2zt在點(diǎn)(x,yz|的切向量為3t2tx2yz6的法向量為n12,1n(3t2,2t,1)(1,2,1)3t24t10t1,t3所以，該點(diǎn)為

27

n(3x0, 求橢球面3x2y2z29x2yz0的切平面方程.解：設(shè)曲面3x2y2z29在點(diǎn)(n(3x0, z3x0y0 3x0

z0t

t,

2t,

t，又3x2y2z29

t2 4tt916t27t

3 x

3,

323,323,0 3(x 3) 3(y 3) 3(z

3) 43或 3(x 3) 3(y 3) 3(z 3)43 3即x2yz 0或x2yz 3xyz試證曲 xyzxyzn(1, 證明：設(shè)P(x,y,z)為曲xyzn(1, xyz xyzz z

(Xx)

(Yy)

(Zz) xyzxyz X Y Z xyzxyzx y

z

zxy 求曲線y22mxz2mx在點(diǎn)(xyz處的切線和法平面方xy 0解：曲線y22mxz2mx在點(diǎn)0

,

,

處的切向量為1,m1 xx0y0yy0)2z0(zz0 xxmyy1(zz0xmy1zxm

第七 方向?qū)?shù)與梯 設(shè)z

f(xyP(xy的某鄰域內(nèi)有定義，l是任一非零向量，el(ab)f(xyPlf

f(xat,ybt)f(x,tf表示函數(shù)f(xyP處沿方向l若f(xyP(xy處可微，則對(duì)任一單位向量el(abf

fx(x,y)af

(xy)b（此也為方向?qū)?shù)存在的充分條件 設(shè)f(x,y)C(1)，則梯度gradf(x,y)為下式定義的向量gradf(xy（或f(x,y）fx(xy),fy(xy

ff(x,y) 梯度是這樣的一個(gè)向量，其方向?yàn)閒(x,yP(x,y)處增長(zhǎng)率最大的一個(gè)方向；其模習(xí)題8—3M0處沿指定方向l3zx2y2

M0(1,

l為從點(diǎn)（1，2）到點(diǎn)

）解：方向l為l(1,3)2(1 3)，而z

2,z

(1,

y(1,所以z

z cosz cos214

31l(1, x(1,3y3

y(1, ux z

M0 l 3(3 3, 3 u z cosz cosz l

x

y

zu

u

,u

z2

z2所以u(píng) cos1cos1cos 3l

zlnxyy24x上點(diǎn)（1，2）解：拋物線y24x在點(diǎn)(1,2)處的切向量為l=(1,2x) 2(2 2 u

z cosz cos 2 2 l

x

y

3 3 求函數(shù)uxy2z3xyz在點(diǎn)(1,1,2處沿方向角為, 數(shù)解：u z cosz cosz l

x

y

z(y2yz) cos(2xyxz) cos(3z2xy) cos111

AD方向的方向?qū)?shù).

f(xyA5解：AB(2,0)2(1,0),AC(0,4)4(0,1),AD(5,12) 5f(f(x,y)|A|cosf|cosf| y

xf(x,y)|A|f(x,y)|A

y

y所以f(x,y)

|cos

|cos35261225

x

y

f(xyz)x22y23z2xy3x2y6zgradf(0,0,0)gradfgradf(0002xy34yx26z6|(0,0,032問(wèn)函數(shù)uxy2zP(1,1,2處沿什么方向的方向?qū)?shù)最大？并求此方向?qū)?shù)的最大值gradu(1,2,2)(u,u,u)

(y2z,2xyz,xy2)

64646464

l第八節(jié)多元函數(shù)的極值及其求必要條件 若f(x,y)在點(diǎn)(x0,y0)有極值且可偏導(dǎo)，fx(x0,y0)fy(x0,y0)0使偏導(dǎo)數(shù)等于零的點(diǎn)(x0y0稱(chēng)為f的駐點(diǎn)（或穩(wěn)定點(diǎn)）.駐點(diǎn)與不可偏導(dǎo)點(diǎn)都是可疑極值點(diǎn)， 充分條件 設(shè)zf(x,y)在區(qū)域D內(nèi)是C(2)類(lèi)函數(shù)，駐點(diǎn)(x,y)D Afxx(x0,y0),B

fxy(x0,y0),C

fyy(x0,y0) （1）當(dāng)ACB20時(shí)，f(xy)A00 （2）當(dāng)0時(shí)，f(x0y0

當(dāng)0f(xyD上的全部可疑極值點(diǎn)（設(shè)為有限個(gè)D邊界上f的最大.最小值進(jìn)行比較，其中最大、最小者即為fDD內(nèi)f(x,y的可疑極值點(diǎn)唯一時(shí)，無(wú)須判別，可直接下結(jié)論：該點(diǎn)的函數(shù)值即為fDzf(xy在約束方程(xy)0L(x,y,)f(x,y)(x,y)Lx0,Ly0,L0，則可求得可疑極值點(diǎn)(x0,y0習(xí)題8—（1）f(x,y)e2x(xy22f(x,y)2e2x(xy22y)e2xe2x(2x2y24y1)

xf(x,

e2x(2y2)

y 2f(x,A

(4x4

f(x,28y3)e,B f(x,2

(y1)2f(x,C

2e，B2AC2e20,Aef(xy在(11f(11e(1121 （2）f(x,y)3x2yy33x23y2f(x,y)6xy6x y

x(y1)

x

x1,f(x,

xy2y y 3x23y26y 可疑極值點(diǎn)有四個(gè)，即O(00A(022f(x,y)

6y

2f(x,

2f(x,

6y點(diǎn)O(0,A-600B006-C-600B2--f(0,0)2,f(0,2)8122（1）f(x,y)2x x24y2解：令F(xyf(xyFx(x,y,)2 18F(x,y

x2 817

17

f 7

17)817

17 （2）f(x,y,z) x22y23z2解：令F(xyzxyz(x22y2Fx(x,y,z,)yzF(x,y,z,)xzF(x,y,z,)xy23最大值f(x,y,z) ，最小值f(x,y,z)2233從斜邊之長(zhǎng)為l的一切直角三角形中，求有最大周長(zhǎng)的直角三角形.F(x,y)xyl(l2x2y2Fx(x,y,)12xFy(x,y,)12yF(x,y,)x2y2lxy

2l2sxyl

zx22y2z62x2y2xoy面距離最小值z(mì)x22y2z62x2y2P(xyzmind|zs.t.zx22z62x2F(xyzuz2(zFx(x,y,z,,u)2x4uxF(x,y,z,,u)4y2uy

(2u)x(2u)y F(x,y,z,,u)2zu F(x,y,z,,u)2zu F(x,y,z,,u)zx22y2 F(x,y,z,,u)zx22y2F(x,y,z,,u)z2x2y26 F(x,y,z,,u)z2x2y26(2u)y2zu

z

y2z3u

y

3y3

x2

z 當(dāng)2u

z

與z62x2y(2u)xz2zz

zz

x2222，當(dāng)2u0yz當(dāng)2u

z

與z2x2y26x

2y0z2P(xyzxoyyx2xy20之間的最短距離yx2P(xyxy20的距離為d|xy22|xy22F(xyxy2)2yx2Fx(x,y,)2(xy2)2x

xF(x,y,)2(xy2)02 2F(x,y,)yx2

y 1P(,)2

xy20的距離為d為最小，且d77xyzxyz50xyyzzxVF(xyzuxyz(xyz50u(xyyzzxFx(x,y,z,,u)yzu(yz)F(x,y,z,,u)xzu(xz)

(yx)(zu)(zy)(xu)y Fz(x,y,z,,u)xyu(xy)F(x,y,z,,u)xyz50F(x,y,z,,u)xyyzzx750yx時(shí)

xyu(xy)xyz50xyyzzx750(zy)(xu)

(zy)(xu)

(zy)(xu) x22uxx22ux2xz50

x22uxz50

z50

x10010 50xy505

所以當(dāng)

時(shí)，Vz501010)25010)250(35010V(50510)2501010250(10

10)2zx2y2xyz1截成一橢圓，求原點(diǎn)到這橢圓的最長(zhǎng)與最短距離zx2解：曲線xyz1P(xyz到坐標(biāo)原點(diǎn)的距離為dx2y2x2y2d s.t.xyzF(xyzu)x2y2z2(x2y2zu(xyzFx(x,y,z,,u)2x2xuF(x,y,z,,u)2y2yu

(1)(xy)2y2yu Fz(x,y,z,,u)2zuF(x,y,z,,u)x2y2z

2zux2y2z F(x,y,z,,u)xyz1 u z當(dāng)1

，所以1xy1x2y2 xyz12x22x1

2x22x10x 2 42 41xy2x2x2y2

3,z3，z3，z 3 3)2 51設(shè)有一平面薄板(不計(jì)其厚度)占有xOy面上的閉區(qū)域D薄板上分布有密度為(xy)的電荷且(xy)D上連續(xù)試用二重積分表達(dá)該板上全部電荷Q解板上的全部電荷應(yīng)等于電荷的面密度(xy)在該板所占閉區(qū)域D上的二重積分Q(x,y)dD2設(shè)I1(x2y2)3dD1{(xy)|1x1又I2(x2y2)3dD2{(xy)|0x1 I1表示由曲面z(x2y2)3與平面x1y2以及z0圍成V的體積I2z(x2y2)3x0x1y0y2z0圍成的V1的體積顯然V關(guān)于yOz面、xOz面對(duì)稱(chēng)因此V1是V位于第一卦限中的部分故V4V13利用二重積分的定義證明(1)dD

證 n 0f(xy)1f()1所以ndlimilimn 0

kf(x,y)dkf(x,

(k為常數(shù) 證明kf(xy)dlimkf(i,i)ilimkf(i,i 0n

klimf(i,i)ikf(x,y)d0 f(x,y)df(x,y)df(x,y)d 證明將D1D2n1n2個(gè)小閉區(qū)域i和i n1n2n

i2令各i和i的直徑中最大值分別為1和2 max(12) 12120

10i

20i f(x,y)df(x,y)df(x,y)d 4根據(jù)二重積分的性質(zhì)比較下列積分大小(xy)2d與(xy)3dDx軸y xy1所圍成 區(qū)域D為D{(xy)|0x0yxy1}因此當(dāng)(xy)D時(shí)(xy)3(xy)2(xy)3d(xy)2d (xy)2d與(xy)3dD 解區(qū)域D如圖所示Dxy1的上方(xy)D時(shí)xy1從而(xy)3(xy)2(xy)2d(xy)3d ln(xy)d與(xy)3dD是三角形閉區(qū)域 頂點(diǎn)分別為(10)11)2 區(qū)域D如圖所示顯然當(dāng)(xy)D時(shí)1xy2從而0ln(xy)1故有[ln(xy)]2因 [ln(xy)]2dln(xy)d ln(xy)d與(xy)3dD{(xy)|3x5 區(qū)域D如圖所示顯然D位于直線xye的上方故當(dāng)D時(shí)xye因 ln(xy)d[ln(xy)]2d 5利用二重積分的性質(zhì)估計(jì)下列積分的值Ixy(xy)dD{(xy)|0x1D 因?yàn)樵趨^(qū)域D上0x10y1所以0xy10xy2于 0dxy(xy)d2d 0xy(xy)d2DIsin2xsin2ydD{(xy)|0xD 因?yàn)?sin2x10sin2y1所以0sin2xsin2y1于0dsin2xsin2yd1d 0sin2xsin2yd2DI(xy1)dD{(xy)|0x1D 因?yàn)樵趨^(qū)域D上0x10y2所以1xy14于d(xy1)d4d 2(xy1)d8DI(x24y29)dD{(xy)|x2y2D 在D上因?yàn)?x2y24所以于 9d(x24y29)d25d 922(x24y29)d2522D 36(x24y29)d100D1計(jì)算下列二重積分(1)(x2y2)dD{(xy)||x|1DD1x11y1(x2y2)d1dx1(x2y2)dy1[x2y1y3]1dx D

1(2x21)dx[2x32x]18 (3x2y)dDxy2所圍成的閉區(qū)域D2D0x20y2x2

22D

[3xy0

]02(42x2x2)dx[4xx22x3]220 (x33x2yy2)dD{(xy)|0x1D

1 3 D

yy

dy11 11

yy

404

yyx]011yy3)dy[yy2y4]111110(4 4 xcos(xy)dD是頂點(diǎn)分別為(00)0)和()的三角形閉區(qū)域DD0x0yx于是xxxcos(xy)d y)dyx[sin(x0 0 Dx(sin2xsinx)dxxd(1cos2xcos x(1cos2xcosx)|(1cos2xcosx)dx3 0 2畫(huà)出積分區(qū)域并計(jì)算下列二重積分D

ydDy

xyx2所圍成的閉區(qū)域1D{(xy)|1

x2y

}xyd x

ydy

2

x1 2x4)dx6x D

x[y2]x2

(x0 xy2dDx2y24y 44D{(xy)|

0x

}4xy2d2 xy2dx2[1x2y2]4y4 D

2 2(2y21y4)dy[2y3

64

exydD{(xy)|D解積分區(qū)域圖如D{(xy)|1x0x1yx1}{(xy)|0x1exyd0exdxx1eydy D

0ex[ey]x1dx1

1y1

0e

]x1dy

e

[1e2x1e1x]0[ex1e2x1]1 (x2y2x)dDy2yxy2x軸所圍成的閉區(qū)域DD{(xy)|

yxy}2

21

12(xD

x)d0dyy(x2

y x]y222(19y33y2)dy1320 3如果二重積分f(xy)dxdyf(xy)f1(x)f2(y)的乘積Df(xy)f1(x)f2(y)D{(xy)|axbcyd}證明這個(gè)二重積分等于兩個(gè)單積分的乘積即 f1(x)f2(y)dxdy[aD

b證明f1(x)f2(y)dxdyadxcf1(x)f2(y)dy

aD cf1(x)f2(y)dyf1(x)cf2(y)dy Dd

f2(y)dy]dx由于

f2(y)dy的值是一常數(shù)因而可提到積分號(hào)的外面 f1(x)f2(y)dxdy[aD

4化二重積分If(xy)d為二次積分(D兩個(gè)二次積分)D是yxy24x所圍成的閉區(qū)域解積分區(qū)域如圖所示并且xD{(xy)|0x4,xyx

}D{(x

0y4,1y2xy4 2 所以I0

f(xy)dyI0dyy2f(xy)dx4xx2y2r2(y0)所圍成的閉區(qū)域解積分區(qū)域如圖所示r2r2r2D{(xy)|rr2r2r2r2rDr2r

0yr,

x r2rr0所 Ir2rr0

f(xy)dy或I0

r2yxx2yr2x解積分區(qū)域如圖所示D{(xy)|1x2,1yxxD{(x

1y1,1x2}{(xy)|1y2,yx2y 所 I1dx1f(x,y)dy或I1dy1f(x,y)dx1dyyf(x,y)dx 環(huán)形閉區(qū)域{(xy)|解如圖所示用直線x1和x1可將積分區(qū)域D分成四部分分別記做D1D2D3D4 If(x,y)df(x,y)df(x,y)df(x,2dx11

11dx114444

f(x,f(x,1

y1y1D分成四部分D1D2D3D如圖所示If(x,y)df(x,y)df(x,y)df(x, 442442

f(x,y)dx

f(x,4141

f(x,y)dx

444441

f(x,5f(xy)D上連續(xù)Dyx、yaxb(b>a)圍成的閉區(qū)域 證明adxaf(xy)dyadyyf(xy)dxD{(xy)|axbayx}D{(xy)|ayb 于 D

adxaf(xy)dy或f(xD

因 adxaf(x,y)dyadyyf(x,y)dx6改換下列二次積分的積分次序y

f(x,y)dxD{(xy)|0y10xy}如圖D{(xy)|0x1xy1}所以1dyyf(x,y)dx1dx1f(x,y)dy 20dyy2f(x,y)dxxD{(xy)|0y2y2x2y}如圖xD{(x

xy2

} f(x,

xf(x,y)dy

2

f(x,y)dx1D{(xy)|0y1

x

1y2}如圖D{(xy)|1x10y1x21122

2xf(x,2x

f(x,D{(xy)|1x22xyD{(xy)|0y12yx

2xx2如圖1y2}22

f(x,y)dy

ln

2 f(x,y)dyD{(xy)|1xe0ylnx}如圖D{(xy)|0y1eyxe}所以 f(x,y)dy0dyeyf(x, 0dxsinxf(xy)dy(2D{(xy)|0xsinxysinx如圖2D{(x,y)|1y0,2arcsinyx 所

f(x,y)dx 7Dxy2yxx軸所圍成(xy)x2y2求該薄片的質(zhì)量M(x,

(x2

1dy2y(x2D[1[

D(2y)32y27y3]dy40 8x0y0x1y1z02x3yz6截1解四個(gè)平面所圍成的如圖所求體積1VD

1dx

2xy3y2]1dx192x)dx70

9x0y0xy1z0x2y26z截得解在xOy面上的投影區(qū)域?yàn)镈{(xy)|0x10y1x}所求的體積z6x2y2為頂D為底的曲頂柱體的體積即 V(6x2 (6x2y2)dy17 D10求由曲面zx22y2及z62x2y2所圍成的的體積z62x2z62x2

消去z得x2+2y2=62x2y2即x2y2=2故在xOy面x2y22xy軸均對(duì)稱(chēng)x都是偶函數(shù)V[(62x2y2)(x22y2)]d(63x22D2

2x2(2x2y2)dy 0

(2x2)3dx611畫(huà)出積分區(qū)域把積分f(xy)dxdy表示為極坐標(biāo)形式的二次積分DD是{(xy)|D如圖D{()|020a} 2daf(cos,sin)d {(xD如圖D{(,)|02cos 2

f(cos,sin)d{(xy)|a2x2y2b2}D如圖D{()|02ab} 2dbf(cos,sin)d {(xy)|0y1x解積分區(qū)域D如圖因?yàn)镈{(,)|0,0 }所2

cos 2dcossinf(cos,sin)d 12化下列二次積分為極坐標(biāo)形式的二次積分 D如圖所示D{(,)|0,0sec}{(,)|,0csc} 所 D

4

f(cos,sin)d24

f(cos,sin)d2 f2

x2y2)dy D如圖所示D{(,)|,02sec} 2所 f(x2y2)dyf(x2y2)df2 34

f()d11

D如圖所示D{(,)|0, 2cos所

2d f(cos,sin)d0

D如圖所示D{(,)|0,sectansec}4 所 0dx0f(x,y)dyf(x,y)df(cos,sin 4d f(cos,sin 13把下列積分化為極坐標(biāo)形式并計(jì)算積分值

2axx2(x2y2)dy D如圖所示D{(,)|002acos2 (x2y2)dy2

2

2

2cos0

4a4

x2y2dy D如圖所示D{(,)|00asec4 x2y2dy

a3 4d

d 4sec3d

[2

2

3 2dx2

y

D如圖所示D{(,)|00sectan4

2 dx2(xy)2dy2 D 4d 2d4sectand

aa2a (x2y2)dx D如圖所示D{(,)|00a2a2a2

2

a2

4 y d 2d D14利用極坐標(biāo)計(jì)算下列各題ex2y2dDx2y24所圍成的閉區(qū)域DD{()|0202}ex2y2de2 ln(1x2y2)dDx2y21D內(nèi)的閉區(qū)域D{(,)|0012ln(1x2y2)dln(1 2 20d ) 22(2ln2 4(2ln22arctanydDx2y24x2y21y0yx 一象限內(nèi)的閉區(qū)域D{(,)|0,124yy

4dd4d

d

15選用適當(dāng)?shù)淖鴺?biāo)計(jì)算下列各題yyxdxdyDx2，yxxy1所圍成的閉區(qū)域DD{(xy)|1x2,1yxxx 2 x yy2dxdy1yyD

2dy1x

4D的閉區(qū)域

1x2y2d 其中D是由圓周x2y21及坐標(biāo)軸所圍成的在第一象限1x2D{(,)|001221x2

11

2d

dd2d

2d

(2)1x

01 (x2y2)dDyxyxayay3a(a>0)所圍成的閉區(qū)域DD{(xy)|ay3ayaxy}(x2

3a

(x2y2)dx3a(2ay2a2y1a3)dy14a4 DD

x2y2dD是圓環(huán)形閉區(qū)域{(xy)|D{()|02ab} 3x2y2ddr2dr2(b3a3)3 D16D由螺線2上一段弧02

)與直線2圍成它的面密度為(xy)x2y2求這薄片的質(zhì)量在極坐標(biāo)下D{(,)|0022

M(x,y)d2dd42d D17y0ykx(k>0)z0R的上半球面所圍成解此在xOy面上的投影區(qū)域D{(x VD

R2x2y2dxdyarctankd

R22d1R3arctank318xOyx2y2ax圍成的閉區(qū)域?yàn)榈讂x2y2為頂xOyD{(xD{(,)|0acos

a4

V

y)dxdy

2

d

2cosd

32a44

x2y2 1化三重積分If(xyz)dxdydz為三次積分其中積分區(qū)域分別是xyzxy10z0所圍成的閉區(qū)域解積分區(qū)域可表示為{(xyz)|0zxy0y1x 于

zx2y2{(xyz),|x,2y2z

y

x}1于 I1

2dy

f(x,y,z)dz12 x12{(xy

x2y2z

2

x12于 I2

2dy

f(x,y,z)dz x提示zx22y2z2x2xOy

x2y21z0所圍成的在第一卦限內(nèi)的閉區(qū)域 {(x,y,z)|0zxy,0y

a2x2,0xa于 Ia

bba2dy

f(x,y,z)dz 提示區(qū)域czxy2設(shè)有一物體占有空間閉區(qū)域{(xyz)|0x10y10z1}在點(diǎn)(xy1處的密度為(xyz)xyz計(jì)算該物體的質(zhì)量1111解Mdxdydzdxdy111

yz)dz1dx(xy

1y21y]1dx

3如果三重積分f(xyz)dxdydzf(xyz)f1(x)、f2(y)f3(z)的乘積f(xyz)f1(x)f2(y)f3(z)積分區(qū)域{(xyz)|axbcyd證明這個(gè)三重積分等于三個(gè)單積分的乘積f(x)f(y)f(z)dxdydz

f

f

mf(z)dz

ab

cm

證明f1(xf2(yf3(z)dxdydz[

f1(x)f2(y)b

a

f3(z)dz)dy]dx[(f1(x)

a

b fb

f

mf(z)dza

4計(jì)算xy2z3dxdydz其中zxyyxx1z0閉區(qū)域{(xyz)|0zxy0yx2

x xy

x2z4于

zdxdydz0xdx0y

z

0xdx0y[4]01

y5dy

1

dx1x4 x

28

5計(jì)算 其中為平面x0y0z0xyz1所圍成的四面體(1xy{(xyz)|0z1xy0y1x于 1dx1xdy1xy dz(1xy 1dx1 1]dy1 31x]dx

02(1x)81(ln25)提示

1dx1xdy1xy dz 1dx1 ]1xydy1dx1 1]dy

2(1xyz)2

1 02(1xy)

dxdx

38

0[1ln(1x)3x1 1(ln25)0 6計(jì)算xyzdxdydz其中x2y2z21卦限內(nèi)的閉區(qū)域{(x,y,z)|0z1x2y2,0y1x1x2

1x2,0x于 xyzdxdydz0

0 7計(jì)算xzdxdydz其中是由平面z0zyy1以及拋物柱面yx2閉區(qū)域{(xyz)|0zyx2y11 x2于 xzdxdydz 1 x211x(1x6)dx0x2x28計(jì)算zdxdydz其中z

zh(R0h0) 的閉區(qū)域解當(dāng)0zh時(shí)過(guò)(00z)作平行于xOy面的平面截得的截面為圓x2y2Rz)2DRz面積為R2z2 zdxdydz

dxdyR2hz3dzR2h2

h2 9利用柱面坐標(biāo)計(jì)算下列三重積分2x2zdv其中是由曲面z 及zx2y22x2202012z2 2于 zdv

211(220(1235)d7( (x2y2)dv其中x2y22zz2所圍成的閉區(qū)域于 (x2y2)dv

2z222dddz

0

20

2222d2(2315)d28d16

0 10利用球面坐標(biāo)計(jì)算下列三重積分(x2y2z2)dv其中x2y2z21所圍成的閉區(qū)域020于 (x2y2z2)dv 2dsind1r4dr4 zdv其中閉區(qū)域x2y2(za)2a2x2y2z2所確定02,0,0r2acos4于 zdvrcosr2

4 40

4(2acos)

5748 40

611選用適當(dāng)?shù)淖鴺?biāo)計(jì)算下列三重積分xydv其中x2y21z1z0x0y0限內(nèi)的閉區(qū)域0,01,0z12于 xydvcossin

1

cos0

111111x1

xydv

dz

0(2

x32[x2x4]11 8

x2y2z2dv其中x2y2z2z所圍成的閉區(qū)域02,0,0rcos2于

x2x2y2

d2

4 20

4 (x2y2)dv其中4z225(x2y2)z5所圍成的閉區(qū)域02,02,5z5于 (x2y2)dv

0

20

525223(55)d8 x2y2(x2y2)dvx2y2

Az0所確定02,0,arA2于 (x2y2)dv(r2sin2cos2r2sin2sin2)r22

3A

55 2

ra

15 a12利用三重積分計(jì)算下列由曲面所圍成的的體積x2z6x2y2及x2 60202 60于 Vdvdddz0

22(623)d32 x2y2z22az(a0)x2y2z2(z軸的部分)解在球面坐標(biāo)下積分區(qū)域可表示為02,0,0r2acos4于 Vdvr2

d4

8

33 30

sin x2z x20201于 V

dv2

dz

123)d

5x2z x5x2502,02,1254于 V

55222(522)d2(554) 13R的球體在其上任意一點(diǎn)的密度的大小與這點(diǎn)到球心的距離成正比求這球體的質(zhì)量x2y2度函數(shù)為(x,yx2y2020 于 Mkx2y2z2dvd