高等數(shù)學(xué)統(tǒng)考試卷（20032004學(xué)年第二學(xué)期）參考解答一、1（漏“一”號(hào)扣一分）4,792DYXYX22310DF475Y0EKX二、6D7D8C9B10C三、11解法1記22,FXYZGXYZXVUXG2VUYVUZGYF，ZVUXYZV22Y221VUVUVUYGZYXZGXZXGYYVV24解將原方程兩邊同時(shí)對(duì)X、Y求導(dǎo)ZZX,Y得（1）02ZXZYVU（2）2GY聯(lián)立（1）、（2）消去GU、GV得ZZZZXYXXYY02212設(shè)三條移長(zhǎng)分別為X,Y,Z，則長(zhǎng)方體表面積為求U2XY2ZX2YZ，其中XYZ3A方法一由得ZYXFF11YXZ得XYZA為所求唯一解故當(dāng)XYZA時(shí)U6A2為所求條件最大值方法二作3,ZYXAYZXZYXF解科XYZA（唯一解）02XYZ（一般不要求判定）判定法（亦是初等解法）222116836AUUXYZXYZX203XYZ2AU且等號(hào)僅當(dāng)XYZA時(shí)或立，故XYZA時(shí)U取得條件最大值26UA13記,21N,ZYXZYX令即代入曲面方程,/2所求點(diǎn)為2,1,2922YY1Y或2,1,214原式2AAXDDAA2115方法一（投影法，柱面坐標(biāo)法）原式22RXYXYDDZD2234RDXYY20322RR223/21240R4441258328RR方法二截面法，用平行于XOY平面的平行平面截所給立體域截面積RZZRZSD20221原式DXYZXYRDD220212RZZZ4224316412RRR56158144R15方法（球面坐標(biāo)法）作錐面將分為1及2兩部分3原式12ZDVZ3002023COS2022CSINCOSINRRDDDD324452031ISICO681RR4425632172PQXYUYXUY故積分與路徑無(wú)關(guān)選L1，從點(diǎn)A5,0到B3,4225DXABXDXX1253532348亦可改選L2折線(xiàn)A5,0,C3,0,B3,434225096ABCBXDYYD92598311VU,2YVX18作輔助01Z原式1上下上外上22211066DVXZYXDVZYX5220202SINR5521820COS222RDRDYXRVDX3/COS2001|RR332SIND191|1UIMLXUILNN當(dāng)|X|原級(jí)數(shù)發(fā)散當(dāng)X1當(dāng)1時(shí)原級(jí)數(shù)收斂UN當(dāng)時(shí)原級(jí)數(shù)發(fā)散當(dāng)X1當(dāng)1時(shí)原級(jí)數(shù)絕對(duì)收斂1NN當(dāng)0時(shí)原級(jí)數(shù)條件收斂當(dāng)原級(jí)數(shù)發(fā)散20記0NB1NNLIMLIE故RE當(dāng)冪級(jí)數(shù)絕對(duì)收斂X|23|1當(dāng)冪級(jí)數(shù)發(fā)散EX322121XY解標(biāo)準(zhǔn)化122XEYDX方法一先解求得021XCY改設(shè)代入方程（）1YU22XEX2XEUCE故得2212XEY方法二P21XPXD2LNL21XEPDX1PXDECY2212XE方法三原方程為221XEYCEYXX2221C22先解065Y由得2R3,21R故知231XXYCE再求的特解，AEY65Y