上海交通大学学年高等数学高数期末考试解答

解B2ACf2(ab)f(ab)f(ab)0，排除A、B.xy xx yyf(x,b)在点xa处取得极小值：f(a,b)0，同理：f(a,b)0. xx yy答案：C解WFurdrr[yzx(t)xzy(t)zz(t)]dt C 0[tsin2ttcos2tt]dt2tdt2 0 0答案：B解S:z1(x2y21)，方向为下侧，IÒ[2y2y1]dvdxdy SS S513123答案：A解|(1)na|1，发散――A错nnn n1 n1 |aa| 1 1 ，发散――B错 n1nn1 n1n(n1)n1(n1)2|aa||(1)n1(1)n|2，发散――C错n1n1nn1(n1)nn1(n1) |a a||(1)n1(1)n|(n1)n n1n1 n n1(n1) n n1 n(n1)11，收敛――D对n1n(n1)((n1)n)n1n2n答案：DS(0)S(0)解S(3)S()02答案：D解1D{(r,)|r2cos2}，xy2dxdy.......D解2xy2dxdy*dy**xy2dx0 * **D557、解蜒2x2xy3y2ds2x23y2ds?x2y2ds25c c 2c 2 P Q8、2xexcosy y x解1(2xyexsiny)dx(x2excosy)dy0(2xydxx2dy)(exsinydxexcosydy)0d(x2y)d(exsiny)0通解为：x2yexsinyC解2u(x,y)(2xyexsiny)dx(x2excosy)dy(0,0)y(x2excosy)dyx2yexsiny0通解为：x2yexsinyC5(5(2)(3)3zxyzzxyzxyxyurur9、div(rotF)(F)0 x yyz2xz10、解a(x1)n的收敛半径R2nn1na(x1)n1(n1)a(x1)n的收敛半径R2，n n1n1 n0(n1)axn的收敛半径R2n1n1u z11、3e3xyz22e3xyz x x2yz 3e3xyz22e3xyz( )3ezxy u 21 2 32( )3e x 3e1 3(0,1,1)解I1dy2yexydx1 12 y1(e2ye)dy1(e22e)122解C:y0（x:15），1ÑCC1C1x1 x2y2 y 5xdx [(2y ) ]dxdyxx2y2 x2y2 15xdx2122512122dxdy 1 2 2D11xy解1(1)SdS1y211xy x z z S Dxz(2)SdS1x2x2dydz y z S Dyz SdS1(2y)202dydz O S D zyz1（D：z0,zy和y1所围成的三角形区域）yz1dyy14y2dz 0 01y1y1y14y2dy551012解2C:yx(0x1) z212xSzdsyds1212x O 1 x114xdx551012合一投影法：PdydzQdzdxRdxdy(PcosQcosRcos)dSP,Q,Rcos,cos,cosdSP,Q,Rndxdyv Dxyv其中zz(x,y),nz,z,1 x y解1合一投影法：原式3x,y,2z2x,2y,1dxdyx2y22y(6x22y22z)dxdyx2(y1)218x2dxdy8u2dudv4(u2v2)dudv x2(y1)21 u2v21 u2v2114224解2Gauss公式设:z2y(x2y2z)，取上侧，则原式Ò SS 312dV3xdydzydzdx2zdxdy zdxdz4ydxdy2 z2 x2y22yx2z4z2(1)dxdz[4(y1)4]dxdy2z2 x2( )21 x2(y1)2122(v1)dudv4[v1]dudvu2v21 u2v212dudv2u2v21解对级数(1)n3n1yn，un1 2n1 u n0 n 2n3 133，R, 2n1 3y1时，(1)n3n1(1)n3发散，3 2n1 3 2n1n0 n01y时，(1)n3n1(1)n(1)n3收敛，3 2n13 2n1n0 n0得(1)n3n1yn的收敛域为：(1,1]， 2n1 33n02x211故原级数的收敛域为：3,3，2x2即x2,11,2. 1nn (n1)11解n132nn1n1(9)nn1n119nn1n1191n 11xn| 10n1n1 x19Sx1xn11xn11x(xn)dxn1 x n1 x0 n1 n1 n11x(x)dx1[xln(1x)]x01x xn11nn1S191019(19ln109)1099ln109 n132n 10 证(1)a2,aaa2a22.假设a2n2，34323n则aaa2a2n1,故n3:a2n2.n1nn1n n1 n1,(2)axn12n2xn12xn2故当x时，级数axn1（绝对）收敛.nn1S(x)