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2008级《高等数学》第二学期期末考试解答(180 A卷)
一、单项选择题(每小题3分,共15分) 1. 设u(x,y)?(A)
?u?x2(x?y)?x?y???u?y222x?yx?y?(t)dt,其中:??u?x22(t)具有一阶导数,则(
2 )
22?; (B)
???u?y22;
(C)解
?u?x2??u?x?y2; (D)
(x?y?)??u?y22??u?x?y.
) ,
ux?2(x?y)?1??(x?,yuxx?2???(x?y?)??x(?yuy?2(x?y)?1??(x?y?)?(x?,yuyy?2???(x?y?)??x(?y) ,
答案:A 2. 函数z?xy(3?x?y)的极值点是
( )
(A)(0,0); (B)(1,1); (C)(3,0); (D)(0,3). 解1
zx?3y?2xy?y2,zy?3x?2xy?x2,A、B、C、D都是驻点,
zxx??2y,zxy?3?2x?2y,zyy??2x2,
AC?B2?4xy?(3?2x?2y)?0,仅当(1,1)满足
答案:B
解2
x,y对称,C对,D也对,单选题,故排除C,D,
,(x,y)?(0,0),z?3xyz?xy(3?x?y)?3xy可正可负,不是极值点,
答案:B 3. 设有空间区域?1
?2:x2:x?y?z?R,2222?R,z?02与
( )
;
?y2?z2x?0,y?0,z,?0则
(A)????1xdV?4???xdV?2; (B)????1ydV?4???ydV?2 (C)???zdV?1?4???zdV?2; (D)????1xyzdV?4???xyzdV?2.
解 答案:C
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4. 一个形如?bnsinnx的级数,其和函数S(x)在(0,?)上的表达式为
n?1?12(??x),
则S(x)在x?3?处的值S(3?)= ( )
22(A)
?; (B)??44; (C)
?2; (D)??2.
解 S(3?)?S(3??)?S(???1122?22)??S(2)??2(??12?)??4?
答案:B ?5. 若级数?(?1)na的取值范围是 ( n?2na?(?1)n收敛,则(A)a?0; (B)a?1; (C)a?132; (D)a?1.
?(?1)n??(?1)n[na?(?1)n]?解 ?(?1n)na?n?2na?(?1)n?n?2[na?(?1)n][na?(?1)n?]?n?2n2a?1
1?(?1)nna?a???(?1)nn?0时收敛,
n?2n2a?1n?2n2a,a?1???12a?1,即a?1n?2n2a,?12时收敛,
答案:C
二、填空题(每小题3分,共15分) 6. 设D?{(x,y)|x|?|y|?1},则二重积分??(x?|y|)dxdy?__________
D解 ??(x?|y|d)xdy???4ydxdy
DD1?4?11?y10dy?0ydx?4?y(1?y)dy?203
7.
向量场????A?(x2?2y)i?(y2?2z)j?(z2?2x)k,则rotA??__________.
ijk解
ro?tA?????x?y?z?(2,2, 2x2?2yy2?2z2z?2x J-2
)8. 曲面z解
?x?y在点(1,9,4)处的切平面方程是:________________.
12x,?21y,1),
?n?(?zx,?zy,1)?(?11?n|(1,9,4)?(?,?,1),或(3,1,?6),
26切平面:3(x?1)?(y?9)?6(z?4)?
9. 设C为球面x2解 ?C20,或 3x?y?6z?12? 0?y?z22?a2与平面x?2y?z?0的交线,则?Cxds2=____
x2ds?13?(Cx?2y?2)z1d?s32?aC1?ds32?2?a23??a3
10. 级数?n?1?x2nn的收敛域为 :___________
解
|x2n2n|?|1?0,x1?n|x|?1n[?1,1 ]|???,x|?|,收敛域为:12?2????,|x|?1三、计算下列各题(第1小题6分,第2小题8分, 共14分) 11. 设zdz?z(x,y)由方程F(2x?3z,2y?z)?0所确定,其中:F是可微函数,求
.
?zxdx?zydy2F1?3F1?F2解1 dz
dx??2F2?3F1?F2dy?2F1dx?2F2dy3F1?F2??
解2 F1?(2dx?3dz)?dz?F2?(2dy?dz)?0
2F1dx?2F2dy3F1?F2
1xxx112dx?112. 求二重积分:?2142edy??ydx?xxedyy.
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x解 I??112dy?yy2edxy
3e8121??112y(e?e)dy?y?e2
四、计算下列各题(每小题10分,共30分) 13. 设曲面?为柱面x2分析: 求柱面x21.用公式: I?z22?1介于平面y?0和x?y?2之间部分,求???zdS.
?z?1部分的面积
22???z1?zx?zydxdyDxy, 用S:z??1?x2求导,?
2.用公式: I???z1?xy?xzdydzDyz22, 用S:x??1?z2求导,
3.不能用公式: I???z1?yx?yzdxdzDxz22,用???求导
解 ?1:z?1?x,?2:z??1?x22,
z Dxy?{(x,y)0?y?2?x,?1?x?1}??zdS???zdS???zdS
??1?2222 1?xdSy ????11?xdS?????2
2 x x1?x2?0 [1?yrx21?zx?zy22?1?(?)2?11?x2]
14. 计算:?C?xr?3dx?3dy,其中C为上半圆周
2y?x?x2,方向从?1,0?到
?0,0?,r解1
?P?y????1?y?. =
?Q?x??32?[x232?[x22x(1?y)52x(1?y)5,
??1?y?2]2??1?y?]22?C?xr3dx?1?yr3dy??(1,0)(0,0)?xr3dx?1?yr3dy
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