微积分下第九章重及其应用

一、（一）f(x,D

n f( 0DXf(x,y)dbdxy2(x)f(x, DD为Yf(x,y)dddyx2(y)f(x, Df(x,y)ddr2()f(rcos,rsin D（二）nf(x,y,z)dvlimf(i,i,in 0f(x,y,z)dvbdxy2(x)dyz2(x,y)f(x,y, z1(x,f(x,y,z)dvf(rcos,rsin, f(x,y,z)dvf(rsincos,rsinsin,rcos)r2sin 2y（1）2 f(x,y)dx2y

f(x,y)dx21 f(x,y21y4y4y

f(x,y)dx 1dx1f(xy)dy1dxx2f(xy)dy y2dyy2dy

y1f(x,y)dxy1f(x,y)dx1f(x,y)dx 1dxx2f(xy)dy2x2ydDy2xy3 x2yd2dx3xx2 12x2y2|3x2 2x12[x2(3x)227yxd，其中D由x ，xy2及x轴所围yD 2xd0dyy

y11x2|2yy211[(2y)22111(x2y2D

]dxdyDyxy1x11(x2y2xyeD

dxdy01 1所以原式D

dx111(x221(x21)dx112 x2y21d，D:x2y24D解 x2y21d2d2r2 D2(1(1r2)rdr2(r2 5（1）(xyz)zdv，:x2y2z2解：由对称性知(xzyz)dv0所以原式z2dv1(x2y2z2 312dd1r4sin3 44x2y（2）zdv，4x2y4r4r1

，x2y21zdv

d0

121(4r2 7x2x2y4x2y（3）(x2y2z2)dv，

z 解：原式

d4d

r4sin

2)1 64(15

2)2解：令

x2y2zdv，:x2y2z2x2y2zx2y，得xoyDx2y21

z2x2yx2x2y

2r

zdv

zr212r2(1r2081x2y4x2y1(x1x2y4x2y1(x2y23x2y2x2y2z

xy0．zMzMMzzdv

d3d

r4sin 2

1sin2|020

1r515193Mdv

d3d

r3sin 02(cos|3)0

r4141154zMM

31（2）Iy(x2z2)dvx2dvy2dv1(x2y2dvI

Ix

x2y2

z

22d3d

sin2cos2)sinr 111112r6|22r6|23sin2cos2)sin2

3(1cos2)sin02

(cos

1cos3)|030133zI(x2y2)dvz

d3d

sin3r5dr223sin31r6|2r6|2

3

0cos3cos)|0358（3）FFxFyFz}FxFy0Fz

G1dvcos，其中r

z，r x2x2y2zFz

xyxy

d3d

rsincosdr

F0,09F89.5f(t在[RRf(z)dvR(R2z2)f(z)dzR2rR其中x2y2z2R2rRf(z)dz

d0

rfR2rR2rRR2rR2rR2rRR2R2z

rf(z)dz 0f R

rfR(R2z2)f(z)dzI(xy)dDx2y1．Dxd0D 1Iyddxx21D11y2|12dx (1x1 1 21(1x4)dx4 Ixy(xy)2dDyx,y1x1DI(x3yy3x2x2y2dD(x3yy3x)d0D2x2y2d8x2y2d D1xy10x1I8x2y2d81dx1x2y2dy81x2y3

dx

1(x2x5)dx43

3 I(x2y2dDx2y22yx2y22xDI

40

20

r2dr

24

20

r2dr38 83 4sin3d3

38 83 4(cos21)dcos3

2(1sin2)d48(4

2)3 3xyI0dxx

dyy13y131I1

ey

dx

0(yy)e11(1y2)ey2dy211(1y22 21[(1y2)ey2|11ey2dy2 e1．4x2yx2y4x2yx2y4r解：I2d1dr14r

z 4r121(1r4r 1r1r4r

dr24411rdr1444424

4rdr31 dr233

210)3求lim

f(x2y2z2dvf(uf(0)0t0t解：

x22z2tff(x2y2z2t0t

x22z2tlim 2

tf(r)r2sin t0t4

t

4tf(r)r2dr0t

t