微积分iii-2003春十五周答案

2003- 2004-5-x2y2z21.曲线x2y2z2

。15答案： 152．设Dx

xy1，则积分x

yd

2R3．I 20

D f(x,y Rdx

3R2fxR2 。答案3dR

f(rcosrsin)rdr R3u(x,yzSnSudSS uxcosdydzuycosdzdxuzcosdxdyudss其中n=cos,cos,cosT已知方程y1xy2xy

fxfx0 ysinx,yx2sinx,ye2xsinx，则此方程的 答 yc1x2c2e2xsinx（注意答案不唯一

t0ft

1cosx2y2d是t x2y2t9.函数yy(x,z)由方程xyzexy所确定，则y y(x yzex2xxx(1yexy2xxfu4fudu4，L0

y 起点为0,终点为2,0则fx2y2xdxydy 答 211.S为平面x2z10含在柱面x2y621S2zxdS .答案y4y4ycos2

5

y0y0

y(C1C2yacos2xbsin

8bcos2x8asin2xcos

b18

a0,所以y1sin8

Yyy

x)e2x+1sin8y(0)0代入Y中，得C10y(0)0代入Y中，得

14于是Yxe2x1sin2x a2x2.设xya2x2.axdydz2xaydz

a0S为2计算2

xy

z

a2（a2

3a2xd ）aIa

axdydz2xaydzx2y2z2

+

axdydz2xay x2y2z2a2x2a2x2y

下侧S2

x2y2a2z a2I axdydza2a2a2

axdydz2(xa)S1a2 (3aa2a1a [3V0a1a a2直接写出下属结果是错误的I axdydza2 x0,f(x为连续可微函数，且f12，对x0的任一闭曲线L，有4x3ydxxfxdy0L

f(x和积分4x3ydxxfxdy0A2,0B2,3PQf(x

f(x)4x2 f(x)

1dx

[C

1dx

dx]

1[Cx4f(1)2代入上式得C1x所以f(x)1x3x 求方程xy"y'3的通解 答案：yc1ln|x|3xc2Iyz)dxzx)dyxy)dzLx2y2z2ayxtan 02 2 Ox

yxtanxcos2

z

axacoscosyasincoszasinI(yz)dx(zx)dy(x

t:0

[(asincostasint)(acossint)(asintacoscost)(asinsin(acoscostasincost)(acost)]dt2a2(cossinIy2z2dxz2x2dyx2y2dzLx2y2z21ABCAA(1,0,0I(y2z2)dx(z2x2)dy(x2y2

0,( )解(

(C

( (y2z2)dx(z2x2)dy(x2y2

L(

L(C)(

LAB

xcosysinz

t:02 L(

(y2z2)dx(z2x2)dy(x2y2)dz 2[sin2t(sint)cos2t(cost)]dt

(CL(

(y2z2)dx(z2x2)dy(x2y2(L(C

(y2z2)dx(z2x2)dy(x2y2)dz3 I(y2z2)dx(z2x2)dy(x2y2)dzI(xyz)2dSSS解:I(xyz)2dS(x2y2z22xy2yz (12xy2yz2zx)dS42(xyyz 其中4是球的表面积.由对称性可知,xydSyzdSzxdS0I求xyzdSSzx2y2S

0z1

xyzdS4xyzdS4xyzdS,其中S1S xyzdS xyzdS xyz1z2z2

x2y2x0,4xy(x2y2x2y2x0,

14(x2y2)dxdy12551x2y2z2R2z0z轴旋转的转动Jz.解 J

(x2y2dSSR2x2R2x2y

(x,y)

(x,y)|x2y2z

zRR2x2yR(x2y2R2x2y

，dS

dxdyR2R2x2y

dxdy

d

d4RRR2RR2x2yR2R3xRsinyRsinsinzR

(02,02 A2B2CA2B2C

Jz

d2R2sin2R2sind0

3求y2zdxdy,其中闭曲面SSzx2y2z1解:y2zdxdyy2zdxdyy2zdxdySSSSSSS侧

侧zdxdyy

(x2

y

上)(dxdy)

x2y2

(x2

y

)(dxdy) 6y2zdxdyy2dxdy，故y2zdxdySDSD

S xdydzSS为旋转抛物面zx2y2zyz平面,侧SS侧而S

S右，右

xdydzxdydzxdySSS SSS侧xdydz

右xdydzxdySSS SSS侧 侧 Syz平面上的投影面积为0xdydzSS11右11SS11侧11

zy2dydz

zy2dz4

zy2(dydz)

zy2dz4侧 xdydz S已知积分(xxysinx)dxf(x)dy与路径无关，f(xf0 f(x对(1)f(x,求函数uu(x,y)使得du(xxysinx)dx

2f(x)dyx对(1)f(xA(,1B(2,0解

f(x)

(xxysinx xsinxxf(x)f(x)，f(x)

f(x)x2sinx f(x)x(sinxxcosxC) f0C1,于是f(xx(sinxxcosx1 2 (xxysinx)dxu

f(x)dy(xxysinx)dx(sinxxcosx1)dyx xxysinx，u

xycosxysinx(2uxcosxsinx(y)sinxxcosx(y)1，(y)yCx2u其中C(x,y

xycosxysinxy2u(0,0)(xxysinx)dx(sinxxcosx1)dy

0xdx0(sinxxcosx1)dyx2xycosxysinxyC2AB

I1(cos1)dyAIu(x,y)Bu(B)u(A)(1)A

xdx(1)2。22, y

y y

1x2cosxdx

X2

cosyx

y22

yY y

y