234234由于（C）f(x,yz)xy的偶函數，面1yozxoz左右對稱，則zdv4 2351F(tz2dvf(x2y2 2dtdhz2dz2dtf(2)dh 1h3t22htf(2)d F'(t)2h3t2hf(t2)t2h3t2hf(t2limF(t)lim

1hhf(t2t0

t 2z

z

，在xoyD:y242x(x0y0z0vxv

dxdyxxDxx D 2 2 2 2 7故形心坐標為(,082362361、 xy x

L:x2y2Ly1I

2yds 2(sin1)(sin)cos

2(sin1)d2

xds0x14x22220 20 (14x2 PAGEPAGE23721L:2

y 1y L2xydsL2xyds(3x4y)ds 22LLxxds2yds z22LLLxds213xyz LLR2 22 xdx2a(tsint)dt3

1ILOA0[(excosya)(excosyb)]dxdy2a0D(ba)a22 xesinydyyesinCesinyesinxdxdyC xesinydyyesinxdxesinyesinCDDyxesinyesinxdxdyesinyesinxdxdy xesinydyyesinxdxesinyesinxd

esinxesinxd2esinxesinxd2d2 2412411、該曲線積分與路徑無關，故取一簡便的路徑，從點(00)到點(x,y u(x,y)x2xdxy(2ycosxx2siny)dy x2cosyy2cosx222

2 2

2

2

2(xy

(x

y

2x

y

(x

y

2y

x2(12)y2(21)

(12)(x2y2)因為yx

12

2y 2u(x,y)12x2x22

(x212)dx1y

(x2y2)2241頁：3、(1,1)xy2dxy(x)dy解：p 2 2xyy'(x)(x)x2 (0)0,c (x) (1,1)xy2dxy(x)dy(1,1)xy2dxyx2dy 1y1dy 2 從

，即2xgy2y22f2 2xf'(y)2y22g'( f'(y)g(故必須有gyy2f

f''(y)g'(y)y2f(

f(y)c1cosy

sinyy2由f(0)

及g(0

c10,c2

f(y)sinyy2

，g(y)cosy2最后，選取點(00經點(0)到點(y0x2I

2[xf(0)g(0)]dx2[2g(y)2y22f( 24241y2z2dx2z2x2dy3x2y2y22z23x2解原式2(2y4z)dydz(2z6x)dzdx(2x2113 33 2的單位法向易為33

1,1

dscos

dscos

dscos23 (4x2y3z)ds2(xy23D12dxdyD2422421(xyyzssxozxydss

yzdss故xzds

x2x2y21(z)2(zDxoy:x2y2z

1(x2y22

2

z

1(x2y22

22代入1z)2z)2dxdy

xx2

2ad

2d2（1）F(xyz)x2y2z2yzpxoy即(Fx',Fy',Fz')(0,0,1) 因Fz'2z故2zyps上，故軌跡c 2zyx2y2z2yz

（2）x2y2z2yz1的兩邊求全微分，得2xdx2ydy2zdzzdyydz0dz

2yzy y 2y從而xy2z yy1(1(z)2(z4(x2y2z2yz)y2z24yzy2z44y2z242z4y2z24從而I4y2z242(x

dxdy由（1）消去z，可得 (x,y)x23y2 1、由高斯可I(2zz2z)dV 2 d4sincosd 3d2 2、由高 可xf(x)dydzxyf(x)dzdxe2xzdxdys[xf'(x)f(x)xf(x)e2x]dVVs的法向量指向外側時，取"號，s的法向量指向內側時，取"sxf'(xf(xxf(xe2x f'(x)(11)f(x)1e2x

x解f(x)

f(x)

e2x1x

lim(e2xcex)

cxxf(xexxx

xdydzxdydzydzdxzdxdy(x2y2z2xdydzydzdx(x2y2z21:x2y2z21故Ixdydzydzdx高 3dxdydz3413 2452454、解

y1(1y

yx2z21 xx2z2S1

y

I[(8y1)4y4y]dxdydz2(1y2 S

先=后

1dydxdz2(13D 3232頁解：設VSh(t則

dxdyh(t)1[h2(t)2 2x2y21[h2(t)h(t)z2h3(t)

16(x2y2 S

dxdy

h2

dxdyx2y2

h2(t

x2y2

h2(t1z1z'2z'

dv0.9S可得dh(t)

h(t)13t 由h(0)130

h(t)13t130h(t)0，得t100（小時247247x2y2x2y2graduu,u,u

div(gradu)xyz div(gradu)x2y2由于uu

1ln(x2y2z2x2x2y2 y2z2, x2