工科数学总练习题六2答案

；x2y2z(yexeyy)dx(xeyex)dyxdydzydzdx(zR)dxdy x2y2z82u2u a2r a2rrdr．a2ra2rx2y2R2（A）x2y2zdSx2y2R2x2y2dxd （B）(x2y2)dxdy(x2y2)dxd （C）zdxdy（D）zdxdy2Rxydxdy[f(t)e]sinydxf(x)cosydyxfLex exe exe exe5(1)5(1)IxdydzydzdxzdxdyS:a2b2c21(外侧 y2设S1zc1x2y2(上侧a x y z－c1a2b2(下侧1dxdy1dxdySSS dxdy c1 c1 由轮换性dydz1dzdx1dxdy4abc(11)2I(8y1)xdydz2(1y)dzdx4y解zyxy1z2xox o *:y3(x2z22)(右侧(8y14y4y)dv1zxD 120(2rr)dr2(132)dzdx （3）[(xy)2z22yz]dS是球面x2y2z22x2zx3dydz(y3f(yz))dzdx(z3f(yz))dxdf(t)是圆锥面xz2y2与球面x2y2z21所围立体表面取外提示：用Gauss注意：f(x)为奇函数，f(x)x3dydz(y3f(yz))dzdx(z3f(yz))dxd=(3x2+3y2+zf(yz)3z2yf(3(x2+y2z26a2b2c21上第一卦限的点 xyz 0,0,由轮换对称性，xdydz=ydzdxI3zdxdy3 y D{(x,y 1,x0,yxb2) x a )dxcb下面求在条 2 (0,0,0)下的最大值令F(,,,)(1 由F0,F0,F0,得222 ac从而有222 2221 由问题的实际意义知, , 33zz解：Q=vndSxzdydz2xydzdxSS添加S:z0,(x,y)D:x zdV3xydxdyzdV0zdz2z(1z)dz解法1解法1用解记为平面xyz2L所围部分的上侧1 1 1Id2(4x2y3x:xyz2,(x,y)DD:xy2(xy6)12Dy1 1DD的形xyI (yz)dx(2zx)dy(3xy [y(2xy)]dx[2(2xy)x (3x2y2)d(2x (122x2y)dxdy12dxdy分I(3xyx)dxf(xy)d和If(xy)dx3xy2x3)dy解：由积分与路径无关知：f6xy,f3y2f6xyxfxy3x2yCyf3x2Cy3y23x2知Cy即Cyy3并并求 2xy(x4y2)dxx2(x4y2)选取，使2xyx4y2dxx2x4y2 令P2xy(x4y2),Qx2(x4y2 x4x(x4y2)(1)2xy(xy)dxx(xy) 2 2 2（x,y）2xydxx 2x0dx xx4 1x4 0x4vysinx2dxxcosy2dy2证明：记证明：记Dx2y2xy0，即(x1)2(y1)21Green， 2又由于具有轮换对称性，所以cosy2dxdycosx2dxdyI(sinx2cosx2)dxdy2sin(x2根据(x1)2(y1)21可知(x1)21 2 2有1(21)x1(2因此x21x32 4故2sin(x21.又D的面积为，可得22I2 2 12．求指数，使得曲线积分ks,trdx2r与路线无关r2x2y2解P:(x2y2)2Q:(x2y2)Q(xy 22x32xy2xxy2xy)P y 22由QP得，2x32xy2x3x3xy2kkrdx rsyx xsxx dy（当x0,y0时 s(s2t2s2t2)ss(1)vxydxxyydvvndsuu uvuv d u vvuux xvunnuucos(n,x)usin(n,x),vvcos(n,x)vsin(n,xn n n=xcos(n,x)xsin(n,G,22 vvLu vudx再用 (xy，求xfyfDxy其中Dx2y2提示：D(rx2y2rxfyfyr rcosfrsinffdyfydx=0D(rfdxdy=redddr= d r0dD(rf(t)e4t f1 xy)dxdy f(x2y24tf(t)e4t rf()dr rf(由(1)f(0)f(t)8te4t8tf(t f(t)8tf(t)8te4t（一阶线性微分方程f(t) [8 dtc](4t 由f(0)1c f(t)(4t16．设对于半空间16．设对于半空间x0内任意光滑有向封闭曲面wxf(x)dzdxxyf(x)dzdxe2xzdxdy设G是空间二维单连通区域，Pxy(PQR0 PQ xf'(x)f(x)xf(x)e2x0x即f'(x11)f(x1e2xx f(ce (1解得f(x) (exc将条件limfx)1代入得cx0f(x)e(ex17．设在上半平面17．设在上半平面D(x,y)|y0内，函数f(x,y)具有连续偏导数，且对任意的t0都有f(tx,ty)tfy).证明：对D内的任意