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一、填空题(每题3分,共15分)1.12.193.84.45.25二、选择题(每题3分,共15分)1.D2C3.B4.A5.A三、(本题满分10分)解:在方程两边关于x求偏导数得1zz(1ezxx(x,y(1,0时,z0,代入上式,得z1z1.类似可得y(1,02x(1,0222zz1zzzzzee(1式两边关于y求偏导数得,代入x1,y0,z0xyxyxyx(1,022z1z1,解得xy(1,08y(1,022zez2z1zz1或者:计算得,同理可得xy(1ez3xy(1,08xy1ez四、(本题满分12分)fx(x,y2x60,解:令得驻点(3,2,(3,2.又2f(x,y3y120,y(x,y2,fxy(x,y0,fxx(x,y6yfyy(3,22,Bfxy(3,20,Cfyy(3,212在驻点(3,2处,AfxxACB2240,故(3,2不是极值点;(3,22,Bfxy(3,20,Cfyy(3,212在驻点(3,2处,AfxxACB2240,且A0,故(3,2是极大值点,且极大值为f(3,218.
五、(本题满分12分)解:记D1{(x,y1x2,2xxyx},则2yyx1OD1Df(x,ydxdyx2ydxdydxD112x2xx2xydy212x(x4x3dx124920六、(本题满分12分)补充曲面1z2(xy4,取上侧.22r21所围成的立体区域,则z2,0r2,02,由Gauss公式可得214zxdydz2zdzdx(1z2dxdy(4z2zdv2drdrr2zdz0022222r432;2r(4dr04324zxdydz2zdzdx(1zdxdy(3dxdy12,1x2y24所以I14zxdydz2zdzdx(1z2dxdy4zxdydz2zdzdx(1z2dxdy13268(1233七、(本题满分12分)求幂级数(3n1xn0n的收敛域及和函数s(x解:limn3n41,所以收敛半径为R1,收敛区间为(1,13n1nnx1时,lim(3n1x0,所以原级数均发散,故收敛域为(1,1s(x(3n1x3(n1x2xnnnn0n0n03(xn12n01x23212xx(1,13(1x1x1x(1x21x(1x2
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