同济高等数学课后习题解答下册

第八 第一 多元函数的基本概

(x,y)(x0,y0

f(x,y)A是点(x,y以任何方式趋于(x0,y0习 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 证明下列函数当(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第二节偏导zf(xy在(x0,y0fxfy

,,

))

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

的几何意义为曲线zf(xy在点M(x,

,f(x,

x

y

0 0f(x,y)在任意点(x,y处的偏导数fx(x,y)、fy(x,yfx(x,y时，只需把yxzf(xy的偏导数fx(xy),fy(xy42z2 2 2x2y2xyyxzxy

习题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)在点(x0,y0处的全增量zx2yzAxByx2y则称z

f(xy)在点(x0,y0)可微，并称AxByAdxBdy为z

f(xy)在点(x0,y0的全微分，记作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的某邻域内可偏导，且偏导数在(x0,y0zf(xy在(x0,y0习 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)

在点(0,0)解:由于limf(x,y)lim(x2y2)

0f(00f(xy在点(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在点(00(1.02)3(1.02)3f(xy)

x3，则dfx3

3x2dx x3再设(x0y012x0.02y(1.02)3 f(xx,yy)f(x,y(1.02)3 13 30.0212(0.03)30.060.3613213x2x6m,y8m解：对角线长为fx2

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

60.0580.1100.5626262第四节多元复合函数的求导设u(x, v(x,y)在(x,y)可偏导，z

f(uv)在相应点有连续偏导数，则zf(xy),(x,y)在(x

zfu

v

zfuf u

v

u

v多个中间变量：设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只有一个中间变量：设u(x

zf(xyuzfxy,(xy)zfuf zfu u u 只有一个自变量：设u(tv(tw(tzf(t),(t),(t)dzfdu

dvf u

v w

习题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 设uf(xxyxyzfxufyfxzf

xf12zxf13fy[xfzxf]zfyz[xf

zxfx2yy)f,有二阶连续导数，求xzfxfy2y1fyf 2z f f

f2x 2z

f(xy,ygxfg有连续二阶偏导数，求zyf1fgyyf

fy y

y 2

fxyfxf

fxf

f

g x1x1

y

y2 y

y3

x311

f2

gyg第五节隐函数的求导公F(xy)0yy(x，则dyFx F(xyz)0zz(xyzFxzFyG(x,y,z)（1）若F(x,yz)0确定yG(x,y,z)

dy

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

，dz

(F,(y,(F,G)(y,（2）若F(x,yuv)0确定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)习题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的一阶偏导数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

设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设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设zlnz xet2dty

求zlnzxet2dt011)dzex2dxey2dy zex2 zex1z,y1

ze (1

x2ze

ye

1

(1

(1

(1设uxy2z3zz(xyx2y2z23xyz所确定的隐函数，求

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

dy,dz

dz

dz2xdx2

dz2ydy

1 dy2xdx4ydy6zdz03zdz2ydy dy

x(16z)dz

x1

,dy

x(16z)2y(13z)

2y(1xeuusin (2)设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

第六节多元函数微分学的几何应 设点M0(x0,y0,z0) 若:xx(t),yy(t), 则切向量为

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)则切向量为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 设点M0(x0,y0,z0) 若:F(x,y,z)0则法向量为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)则法向量为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)则法向量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

对应t1的点处的切线和法平面方程 1 解： ,2t) ,t2(1 yx切线

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

x2y1 x2 y（2） a

，点(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法线： 0 0 0 0 0

c

2 xt3yt2ztx2yz6txt3,yt2zt在点(x,yz|的切向量为3t2tx2yz6的法向量为n12,1n(3t2,2t,1)(1,2,1)3t24t10t1,t3所以，该点为

27

n(3x0, 求椭球面3x2y2z29x2yz0的切平面方程.解：设曲面3x2y2z29在点(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试证曲 xyzxyzn(1, 证明：设P(x,y,z)为曲xyzn(1, xyz xyzz z

(Xx)

(Yy)

(Zz) xyzxyz X Y Z xyzxyzx y

z

zxy 求曲线y22mxz2mx在点(xyz处的切线和法平面方xy 0解：曲线y22mxz2mx在点0

,

,

处的切向量为1,m1 xx0y0yy0)2z0(zz0 xxmyy1(zz0xmy1zxm

第七 方向导数与梯 设z

f(xyP(xy的某邻域内有定义，l是任一非零向量，el(ab)f(xyPlf

f(xat,ybt)f(x,tf表示函数f(xyP处沿方向l若f(xyP(xy处可微，则对任一单位向量el(abf

fx(x,y)af

(xy)b（此也为方向导数存在的充分条件 设f(x,y)C(1)，则梯度gradf(x,y)为下式定义的向量gradf(xy（或f(x,y）fx(xy),fy(xy

ff(x,y) 梯度是这样的一个向量，其方向为f(x,yP(x,y)处增长率最大的一个方向；其模习题8—3M0处沿指定方向l3zx2y2

M0(1,

l为从点（1，2）到点

）解：方向l为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 cos1cos1cos 3l

zlnxyy24x上点（1，2）解：抛物线y24x在点(1,2)处的切向量为l=(1,2x) 2(2 2 u

z cosz cos 2 2 l

x

y

3 3 求函数uxy2z3xyz在点(1,1,2处沿方向角为, 数解：u z cosz cosz l

x

y

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

AD方向的方向导数.

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问函数uxy2zP(1,1,2处沿什么方向的方向导数最大？并求此方向导数的最大值gradu(1,2,2)(u,u,u)

(y2z,2xyz,xy2)

64646464

l第八节多元函数的极值及其求必要条件 若f(x,y)在点(x0,y0)有极值且可偏导，fx(x0,y0)fy(x0,y0)0使偏导数等于零的点(x0y0称为f的驻点（或稳定点）.驻点与不可偏导点都是可疑极值点， 充分条件 设zf(x,y)在区域D内是C(2)类函数，驻点(x,y)D Afxx(x0,y0),B

fxy(x0,y0),C

fyy(x0,y0) （1）当ACB20时，f(xy)A00 （2）当0时，f(x0y0

当0f(xyD上的全部可疑极值点（设为有限个D边界上f的最大.最小值进行比较，其中最大、最小者即为fDD内f(x,y的可疑极值点唯一时，无须判别，可直接下结论：该点的函数值即为fDzf(xy在约束方程(xy)0L(x,y,)f(x,y)(x,y)Lx0,Ly0,L0，则可求得可疑极值点(x0,y0习题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 可疑极值点有四个，即O(00A(022f(x,y)

6y

2f(x,

2f(x,

6y点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从斜边之长为l的一切直角三角形中，求有最大周长的直角三角形.F(x,y)xyl(l2x2y2Fx(x,y,)12xFy(x,y,)12yF(x,y,)x2y2lxy

2l2sxyl

zx22y2z62x2y2xoy面距离最小值zx22y2z62x2y2P(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 当2u

z

与z62x2y(2u)xz2zz

zz

x2222，当2u0yz当2u

z

与z2x2y26x

2y0z2P(xyzxoyyx2xy20之间的最短距离yx2P(xyxy20的距离为d|xy22|xy22F(xyxy2)2yx2Fx(x,y,)2(xy2)2x

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

y 1P(,)2

xy20的距离为d为最小，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时

xyu(xy)xyz50xyyzzx750(zy)(xu)

(zy)(xu)

(zy)(xu) x22uxx22ux2xz50

x22uxz50

z50

x10010 50xy505

所以当

时，Vz501010)25010)250(35010V(50510)2501010250(10

10)2zx2y2xyz1截成一椭圆，求原点到这椭圆的最长与最短距离zx2解：曲线xyz1P(xyz到坐标原点的距离为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当1

，所以1xy1x2y2 xyz12x22x1

2x22x10x 2 42 41xy2x2x2y2

3,z3，z3，z 3 3)2 51设有一平面薄板(不计其厚度)占有xOy面上的闭区域D薄板上分布有密度为(xy)的电荷且(xy)D上连续试用二重积分表达该板上全部电荷Q解板上的全部电荷应等于电荷的面密度(xy)在该板所占闭区域D上的二重积分Q(x,y)dD2设I1(x2y2)3dD1{(xy)|1x1又I2(x2y2)3dD2{(xy)|0x1 I1表示由曲面z(x2y2)3与平面x1y2以及z0围成V的体积I2z(x2y2)3x0x1y0y2z0围成的V1的体积显然V关于yOz面、xOz面对称因此V1是V位于第一卦限中的部分故V4V13利用二重积分的定义证明(1)dD

证 n 0f(xy)1f()1所以ndlimilimn 0

kf(x,y)dkf(x,

(k为常数 证明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 证明将D1D2n1n2个小闭区域i和i n1n2n

i2令各i和i的直径中最大值分别为1和2 max(12) 12120

10i

20i f(x,y)df(x,y)df(x,y)d 4根据二重积分的性质比较下列积分大小(xy)2d与(xy)3dDx轴y xy1所围成 区域D为D{(xy)|0x0yxy1}因此当(xy)D时(xy)3(xy)2(xy)3d(xy)2d (xy)2d与(xy)3dD 解区域D如图所示Dxy1的上方(xy)D时xy1从而(xy)3(xy)2(xy)2d(xy)3d ln(xy)d与(xy)3dD是三角形闭区域 顶点分别为(10)11)2 区域D如图所示显然当(xy)D时1xy2从而0ln(xy)1故有[ln(xy)]2因 [ln(xy)]2dln(xy)d ln(xy)d与(xy)3dD{(xy)|3x5 区域D如图所示显然D位于直线xye的上方故当D时xye因 ln(xy)d[ln(xy)]2d 5利用二重积分的性质估计下列积分的值Ixy(xy)dD{(xy)|0x1D 因为在区域D上0x10y1所以0xy10xy2于 0dxy(xy)d2d 0xy(xy)d2DIsin2xsin2ydD{(xy)|0xD 因为0sin2x10sin2y1所以0sin2xsin2y1于0dsin2xsin2yd1d 0sin2xsin2yd2DI(xy1)dD{(xy)|0x1D 因为在区域D上0x10y2所以1xy14于d(xy1)d4d 2(xy1)d8DI(x24y29)dD{(xy)|x2y2D 在D上因为0x2y24所以于 9d(x24y29)d25d 922(x24y29)d2522D 36(x24y29)d100D1计算下列二重积分(1)(x2y2)dD{(xy)||x|1DD1x11y1(x2y2)d1dx1(x2y2)dy1[x2y1y3]1dx D

1(2x21)dx[2x32x]18 (3x2y)dDxy2所围成的闭区域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是顶点分别为(00)0)和()的三角形闭区域DD0x0yx于是xxxcos(xy)d y)dyx[sin(x0 0 Dx(sin2xsinx)dxxd(1cos2xcos x(1cos2xcosx)|(1cos2xcosx)dx3 0 2画出积分区域并计算下列二重积分D

ydDy

xyx2所围成的闭区域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解积分区域图如D{(xy)|1x0x1yx1}{(xy)|0x1exyd0exdxx1eydy D

0ex[ey]x1dx1

1y1

0e

]x1dy

e

[1e2x1e1x]0[ex1e2x1]1 (x2y2x)dDy2yxy2x轴所围成的闭区域DD{(xy)|

yxy}2

21

12(xD

x)d0dyy(x2

y x]y222(19y33y2)dy1320 3如果二重积分f(xy)dxdyf(xy)f1(x)f2(y)的乘积Df(xy)f1(x)f2(y)D{(xy)|axbcyd}证明这个二重积分等于两个单积分的乘积即 f1(x)f2(y)dxdy[aD

b证明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的值是一常数因而可提到积分号的外面 f1(x)f2(y)dxdy[aD

4化二重积分If(xy)d为二次积分(D两个二次积分)D是yxy24x所围成的闭区域解积分区域如图所示并且xD{(xy)|0x4,xyx

}D{(x

0y4,1y2xy4 2 所以I0

f(xy)dyI0dyy2f(xy)dx4xx2y2r2(y0)所围成的闭区域解积分区域如图所示r2r2r2D{(xy)|rr2r2r2r2rDr2r

0yr,

x r2rr0所 Ir2rr0

f(xy)dy或I0

r2yxx2yr2x解积分区域如图所示D{(xy)|1x2,1yxxD{(x

1y1,1x2}{(xy)|1y2,yx2y 所 I1dx1f(x,y)dy或I1dy1f(x,y)dx1dyyf(x,y)dx 环形闭区域{(xy)|解如图所示用直线x1和x1可将积分区域D分成四部分分别记做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如图所示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上连续Dyx、yaxb(b>a)围成的闭区域 证明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改换下列二次积分的积分次序y

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

xy2

} f(x,

xf(x,y)dy

2

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

x

1y2}如图D{(xy)|1x10y1x21122

2xf(x,2x

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

2xx2如图1y2}22

f(x,y)dy

ln

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

f(x,y)dx 7Dxy2yxx轴所围成(xy)x2y2求该薄片的质量M(x,

(x2

1dy2y(x2D[1[

D(2y)32y27y3]dy40 8x0y0x1y1z02x3yz6截1解四个平面所围成的如图所求体积1VD

1dx

2xy3y2]1dx192x)dx70

9x0y0xy1z0x2y26z截得解在xOy面上的投影区域为D{(xy)|0x10y1x}所求的体积z6x2y2为顶D为底的曲顶柱体的体积即 V(6x2 (6x2y2)dy17 D10求由曲面zx22y2及z62x2y2所围成的的体积z62x2z62x2

消去z得x2+2y2=62x2y2即x2y2=2故在xOy面x2y22xy轴均对称x都是偶函数V[(62x2y2)(x22y2)]d(63x22D2

2x2(2x2y2)dy 0

(2x2)3dx611画出积分区域把积分f(xy)dxdy表示为极坐标形式的二次积分DD是{(xy)|D如图D{()|020a} 2daf(cos,sin)d {(xD如图D{(,)|02cos 2

f(cos,sin)d{(xy)|a2x2y2b2}D如图D{()|02ab} 2dbf(cos,sin)d {(xy)|0y1x解积分区域D如图因为D{(,)|0,0 }所2

cos 2dcossinf(cos,sin)d 12化下列二次积分为极坐标形式的二次积分 D如图所示D{(,)|0,0sec}{(,)|,0csc} 所 D

4

f(cos,sin)d24

f(cos,sin)d2 f2

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

f()d11

D如图所示D{(,)|0, 2cos所

2d f(cos,sin)d0

D如图所示D{(,)|0,sectansec}4 所 0dx0f(x,y)dyf(x,y)df(cos,sin 4d f(cos,sin 13把下列积分化为极坐标形式并计算积分值

2axx2(x2y2)dy D如图所示D{(,)|002acos2 (x2y2)dy2

2

2

2cos0

4a4

x2y2dy D如图所示D{(,)|00asec4 x2y2dy

a3 4d

d 4sec3d

[2

2

3 2dx2

y

D如图所示D{(,)|00sectan4

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

aa2a (x2y2)dx D如图所示D{(,)|00a2a2a2

2

a2

4 y d 2d D14利用极坐标计算下列各题ex2y2dDx2y24所围成的闭区域DD{()|0202}ex2y2de2 ln(1x2y2)dDx2y21D内的闭区域D{(,)|0012ln(1x2y2)dln(1 2 20d ) 22(2ln2 4(2ln22arctanydDx2y24x2y21y0yx 一象限内的闭区域D{(,)|0,124yy

4dd4d

d

15选用适当的坐标计算下列各题yyxdxdyDx2，yxxy1所围成的闭区域DD{(xy)|1x2,1yxxx 2 x yy2dxdy1yyD

2dy1x

4D的闭区域

1x2y2d 其中D是由圆周x2y21及坐标轴所围成的在第一象限1x2D{(,)|001221x2

11

2d

dd2d

2d

(2)1x

01 (x2y2)dDyxyxayay3a(a>0)所围成的闭区域DD{(xy)|ay3ayaxy}(x2

3a

(x2y2)dx3a(2ay2a2y1a3)dy14a4 DD

x2y2dD是圆环形闭区域{(xy)|D{()|02ab} 3x2y2ddr2dr2(b3a3)3 D16D由螺线2上一段弧02

)与直线2围成它的面密度为(xy)x2y2求这薄片的质量在极坐标下D{(,)|0022

M(x,y)d2dd42d D17y0ykx(k>0)z0R的上半球面所围成解此在xOy面上的投影区域D{(x VD

R2x2y2dxdyarctankd

R22d1R3arctank318xOyx2y2ax围成的闭区域为底zx2y2为顶xOyD{(xD{(,)|0acos

a4

V

y)dxdy

2

d

2cosd

32a44

x2y2 1化三重积分If(xyz)dxdydz为三次积分其中积分区域分别是xyzxy10z0所围成的闭区域解积分区域可表示为{(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所围成的在第一卦限内的闭区域 {(x,y,z)|0zxy,0y

a2x2,0xa于 Ia

bba2dy

f(x,y,z)dz 提示区域czxy2设有一物体占有空间闭区域{(xyz)|0x10y10z1}在点(xy1处的密度为(xyz)xyz计算该物体的质量1111解Mdxdydzdxdy111

yz)dz1dx(xy

1y21y]1dx

3如果三重积分f(xyz)dxdydzf(xyz)f1(x)、f2(y)f3(z)的乘积f(xyz)f1(x)f2(y)f3(z)积分区域{(xyz)|axbcyd证明这个三重积分等于三个单积分的乘积f(x)f(y)f(z)dxdydz

f

f

mf(z)dz

ab

cm

证明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计算xy2z3dxdydz其中zxyyxx1z0闭区域{(xyz)|0zxy0yx2

x xy

x2z4于

zdxdydz0xdx0y

z

0xdx0y[4]01

y5dy

1

dx1x4 x

28

5计算 其中为平面x0y0z0xyz1所围成的四面体(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计算xyzdxdydz其中x2y2z21卦限内的闭区域{(x,y,z)|0z1x2y2,0y1x1x2

1x2,0x于 xyzdxdydz0

0 7计算xzdxdydz其中是由平面z0zyy1以及抛物柱面yx2闭区域{(xyz)|0zyx2y11 x2于 xzdxdydz 1 x211x(1x6)dx0x2x28计算zdxdydz其中z

zh(R0h0) 的闭区域解当0zh时过(00z)作平行于xOy面的平面截得的截面为圆x2y2Rz)2DRz面积为R2z2 zdxdydz

dxdyR2hz3dzR2h2

h2 9利用柱面坐标计算下列三重积分2x2zdv其中是由曲面z 及zx2y22x2202012z2 2于 zdv

211(220(1235)d7( (x2y2)dv其中x2y22zz2所围成的闭区域于 (x2y2)dv

2z222dddz

0

20

2222d2(2315)d28d16

0 10利用球面坐标计算下列三重积分(x2y2z2)dv其中x2y2z21所围成的闭区域020于 (x2y2z2)dv 2dsind1r4dr4 zdv其中闭区域x2y2(za)2a2x2y2z2所确定02,0,0r2acos4于 zdvrcosr2

4 40

4(2acos)

5748 40

611选用适当的坐标计算下列三重积分xydv其中x2y21z1z0x0y0限内的闭区域0,01,0z12于 xydvcossin

1

cos0

111111x1

xydv

dz

0(2

x32[x2x4]11 8

x2y2z2dv其中x2y2z2z所围成的闭区域02,0,0rcos2于

x2x2y2

d2

4 20

4 (x2y2)dv其中4z225(x2y2)z5所围成的闭区域02,02,5z5于 (x2y2)dv

0

20

525223(55)d8 x2y2(x2y2)dvx2y2

Az0所确定02,0,arA2于 (x2y2)dv(r2sin2cos2r2sin2sin2)r22

3A

55 2

ra

15 a12利用三重积分计算下列由曲面所围成的的体积x2z6x2y2及x2 60202 60于 Vdvdddz0

22(623)d32 x2y2z22az(a0)x2y2z2(z轴的部分)解在球面坐标下积分区域可表示为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的球体在其上任意一点的密度的大小与这点到球心的距离成正比求这球体的质量x2y2度函数为(x,yx2y2020 于 Mkx2y2z2dvd