微積分三試卷復(fù)習(xí)

浙江大學(xué)2009–2010學(xué)年夏季學(xué)《微積分Ⅲ》課程期末考試試卷注意：解題時(shí)應(yīng)寫出必要的解題過程。（1~914每題6分;10~13每題10分lDxy||x|1,|y|1lD(即對(duì)坐標(biāo)的x2x2y2l

dxy(xyln(x x2y21))d I(5x2y3x4)dx3x5sinydlt,ｔ從０到2I(zy)dxxzdyxy)dL設(shè){(x,y,z)|0z 1xy,x0,y0},求zdv1計(jì)算1

d

d

2(x2(x2y2x2yx2y21xz x21x

1

b2z0,zc所圍成的空間有界閉區(qū)域(abc為正常數(shù)),求z2dv設(shè)z1(x2y2滿足2z8的部分求第一型(即對(duì)面積的)2(xy )SSx2y2z2R2(R0法向量向外，求第二型(即對(duì)坐標(biāo)的)xy2dydzyz2dzdxzx2dxdSIS

x2y2 SSzx2y2滿足0z1的部分z軸交角為銳角求第二型曲(2xz)dydzzdxdy.Sf(u為連續(xù)函數(shù)lA(1,1)yx22x2B(2,2)的有向弧段 lIl

f()dx

[f()1]dy.設(shè)Sx2y2z22z所圍成的有界閉區(qū)域,ncoscoscosS的外法線方向單位向量函數(shù)u(xyz在上具有二階連續(xù)的偏導(dǎo)數(shù)S2u2u2u (ucos S

z,求 y zcos)l為圓周(xx)2yy)2R2正向一周xyR均為常數(shù) x2 x2x2 x2 Il

2x2

: R0; R設(shè)f(x)sin2011x,求f(x)的以2為周期 系數(shù)b2n(n1,2,)解1.參數(shù)法.l:x1cost,ysint,dl x2y2dtdt,0t x2y2dl dt22|cost|dt4costdt 解2.極坐標(biāo).l:r2cos,dl r2r2d2d,2

2x2y2dll

22cos22

82cosd0解3.直角坐標(biāo).上半個(gè)圓:l1:y 2xx2,dl1y2dx dx,0x x2y2dl2 dx22 8 x2y2解 .QPx2y2 3Iy2d41dx1y2dy43

) xx2y2D2.直接法ll1l2l3l4l1y1x11),l2x1y1l3:y1(x:1 l4:x1(y:1x2x21I dxy2dyyln(1 y22)

dx1y2dy1yln(1 y22)dy41y2dy4 解1.加輔助直線段后用.設(shè)AB:y0,x:1lBA負(fù)向D0r1,0 I (15x415x2y2)d 1( l

d

r2cos2r2rdr8151

2cos2d851 12.參數(shù)法.lxcost,ysinttI 0[(5cos2tsin3tcost4)(sint)(3cos5tsin(sint))cost] (5cos2tsin4t3cos6t)dt(cost4)d(cost)sin(sint) 331(525)85426422(5sin4t5sin6t3cos6t)dt82 03.直角坐標(biāo)ly1x2x112I1[5x2(1x2)3x4(3x5 1x2)(1233 2[5x(1x)1 ]dx8xsint1

481.直接法.Lxcostysintz2costsintt0I0[(2cost)(sint)(2cost2sint)cost(costsint)(sintcost)] 2(3cos2tsin2t)dt4(3cos2tsin2t)dt8cos2tdt8 解2.Lx2y21,z2xy,Sz2xy,上側(cè),xyx2y2SIS

dzd

d

2d

SSz

x

xx解3.降維法（LOxy平面的投影曲線Cx2y21,z0,正向I(2xyy)dx(x(2xy))dy(xy)d(2xC(22xy)dx(3x22y)dC1.投影法.Dxyxy1,x0,y0

(3x

1xzdv1dx1x zdz11dx1x(11x 1011

2

1

2(x1)3

22 4

2：截面法.Dxy1z20z111zdv111

dzzdz

1(1z2)2dz1

(z21)2d(z2 4 11(z21)314 x2 z 2x2y2),Dxyx2y2x2原式

x2y2x2y2

4

2

sin

202(cos)4120

(2 截面法.Dz:a2

1

0z z2dv z2dc

z2ab(

1)dzab(

z)c8

S:z1(x2y2),

:4x2y21x2 1x2dS1z2z2dz

d,由對(duì)稱性 (xy)dS0Sz1112S(xy )dS dS x1(x2y2)d12d4r3dr 2 xRIR2S

xy2dydzyz2dzdxzx2dxd利 ,V:x2y2z2R2I (y2z2x2)dV 2

R

sinR2RV1221R54R3

R2

解1.利 ,作輔助平面S1:z1,上側(cè)，xy:x2y2SS1內(nèi)側(cè)，所圍立體:Dzx2y2z0zS(2xz)dydzzdxdyS

1SS1

(21)dvdxdy31zdz31 0 x2.逐個(gè)投影法向yoz面投影,SSS,S:x zy2,cos0,S:xzy2,cos zDz

zy2z1,1z(2xz)dyd

z)dydz

z)dydS1

y

ydzz161(1 dyd

zzzz

y 3

)22ysint16cos4tdt16312 42xoy面投影,S:zx2y2cos0,xyx2y2zdxdy(x2y2)dxdy2d1r3dr211 x (2xzdydzzdxdy11 S1 3.轉(zhuǎn)換投影法xoy面投影,Szx2y2cos0,xyx2y21,n2x,2[(2xzcoszcos]d[(2xz2xx

|cos |cosx[4x2(x2y2)(2x)x2y2x

(x2y2x

12If(y)[yd(1)1dy]1dyf(y)d(y)1d l lf(u連續(xù)F(uF(uf yf(x)d(x lldy

dx2(xln

2(1ln lI2(1ln利

原式x2y2z2)dVz 2 2 226(1cos8)82

QP

2y(y42x2(2x2y4)2,(x,y)(0,0)l不圍繞點(diǎn)(0,0,l圍繞點(diǎn)(0,0取l1xcostysintt

y2dx2xydy2x2

sin3t2cos2tsindtdt2cos2tsin4或取l22x2y41,y2dx2xyd原式 2x2 y2dx2xydy(2y2y)d