一、填空题 1.?3.
1 2. y?C1e?x?C2e3x 620?dy?yf(x,y)dx??dy?yf(x,y)dx 4. 2x?2y?z?2?0
222y425. 1 6. 7. (,2
2442,?) 8.? 9992yy?y19. ?2exdx?exdy 10. ,0
2xx
二、计算题 1. 设
?z?zxz?ln，求 ，
?x?yzy解： F(x,y,z)?xz?ln, zy1yz1xy1x?zFX?,F y???(?2)?,Fz??2????2 ………4/
zzyzyyzzF?zz ……….6/ ??X??xFzx?zFY?zz2 ……….8/ ????yFZy(x?z) 3. 计算
??Dx2d?,D是由直线y?x、x?2及曲线xy?1所围成的闭区域. y2........................................................4?22xxx2解：d???dx?12dy2??1yxyD??212x2x(?)|1dx??(?x?x3)dx...............................................................6?
1yx1192?(?x2?x4)|1?..............................................................................8?244
1
2. 求曲线x?解：xt?1t1?t,y?,z?t2在对应于t?1的点处的切线及法平面方程. 1?tt111?1??,yt?1??2,zt?1?12?1 ……..2/ 1?121xt?|t?1?11?1?|?,y|?|??1,zt?|t?1?2t|t?1?2 tt?12t?12t?14(1?t)t14 T?(,?1,2) ……………..4/
11x?2?y?2?z?1, 即2?y?2?z?1 ……..6/
切线方程为:
1?121?48411法平面方程为: (x?)?(?1)(y?2)?2(z?1)?0,
42x?即2x?8y?16z?1?0 …. .8/
4. 计算
?L(exsiny?y??)dx?(excosy?x)dy，其中L是下半圆周(x?4)2?y2?9逆
时针方向的封闭曲线.
解：设L所围成的闭区域为D,
P(x,y)?exsiny?y??,Q(x,y)?excosy?x
由格林公式
?(eLxsiny?y??)dx?(excosy?x)dy
???(DD?Q?P?)dxdy...............................................................................................4??x?yxx-
???(ecosy?1?(ecosy?1))dxdy...................................................................6?=?2
1dxdy??2??9???9?…………………………….……………………….8/ ??2D 2
?5. 求?xn的收敛域与和函数.
n?1n1解：??lim|n?1|?1,故收敛半径为R?1 …………2/
n??1n?当x?1,?xn???1?n,级数发散, 当x??1,?xn???(?1)n1,级数收敛,
n?1nn?1n?1nn?1n故收敛域为[?1,1) ………………4/
?设s(x)??xnn?1n,则
?s?(x)?(?xn?)???(xn?)???(nxn?1?)??xn?11 .……….6/ n?1nn?1nn?1n?n?11?x故s(x)??x1?x|??ln(1?x) . ..………8
/
01?xdx??ln|156. 求微分方程dy2ydx?x?1?(x?1)2的通解
y?e?25x?1dx(c???2x?1dxdx)......................................................5?解一：
?(x?1)2e3
?(x?1)2(c?23(x?1)2).....................................................................8?解二：对应的齐次方程为
dydx?2yx?1?0,…………….…………………………1/ dyy?2xx?1..............................................................2?
lny?2ln(x?1)?lnCy?C(x?1)2...............4?将C换成u, y?u(x?1)2, ………………………………..5/
dy51则dx?u?(x?1)2?2u(x?1),故u?(x?1)2?(x?1)2,即 u??(x?1)2 3两端积分 u?23(x?1)2?C …..………………7/
3故 y?u(x?1)2?[223(x?1)2?C](x?1) ………………………..8/
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1展成x的幂级数，并求收敛域
x2?x?2111111?1解：f(x)?2??(?)?[?31?xx?x?2(x?2)(x?1)3x?1x?27. 将f(x)?1x2(1?)2]……2/
???1n又 ?(?1)?x??(?1)xn, ?1?x?1 ………4/
1?xn?0n?0
1?12(1???(?1)n(x?)n??(?1)n1x2()n?1xn, ?2?x?2 n?02n?022)f(x)?1?113[1?x?]2(1?x2)
?1?[?(?1)xn???(?1)n(1)n?1xn]?1?1n?023?[(?1)?(?1)n()n?13]xnn?0n?02(?1?x?1)
……….6/ ………..8/ 4
8 计算I?222222，其中Σ是x?y?z(0?z?a)的外侧. xdydz?ydzdx?zdxdy???解：添加平面?1:z?a(取上侧), ?1和?构成一个封闭曲面,设所围的空间闭区域为?,
利用高斯公式
222xdydz?ydzdx?zdxdy?2???(x?y?z)dv.....................................................3???????1?2???xdv?2???ydv?2???zdv????0?0?2?zdz0ax2?y2?z2..................................................................................5???dxdy..........a1?2?z?z2dz??a4...................................................................................................6?02或?2?d??d??(?cos???sin??z)?dz002?aa?2?a1?2?d??[z?2(cos??sin?)?z2?)|a?d?0022?a11?2?d??(a?2(cos??sin?)?a2???3(cos??sin?)??3)d?00222?1111?2?(a4(cos??sin?)?a4?a4(cos??sin?)?a4]d?034481??a421(或?2?dz?d??(?cos???sin??z)?dz??a40002a2?z1或? 2??dxdy?22(x?y?z)dz?0???(a?x?y)dxdy??a4)x?y2DXYDXYa222
又
22224xdydz?ydzdx?zdxdy?adxdy??a ?????1?1故
I???x2dydz?y2dzdx?z2dxdy=
?2???1??xdydz?y2dzdx?z2dxdy???x2dydz?y2dzdx?z2dxdy
?1=
141?a-?a4=??a4 ……………………….8/ 225
三、确定常数?，使得在右半平面x?0上，
?2xy(x4?y2)?dx?x2(x4?y2)?dy
L与积分路径无关.
解：P(x,y)?2xy(x4?y2)?,Q(x,y)??x2(x4?y2)?……………………………..1/
?P?2x(x4?y2)??4xy2?(x4?y2)??1 ?y?Q??2x(x4?y2)??4x5?(x4?y2)??1 ………………………….…3/?X 因为
?L2xy(x4?y2)?dx?x2(x4?y2)?dy与积分路径无关,故
?P?y??Q?x ………………………….5/ 2x(x4?y2)??4xy2?(x4?y2)??1??2x(x4?y2)??4x5?(x4?y2)??14x(x4?y2)???x(x4?y2)??1(4y2??4x4?)
4(x4?y2)??(4y2??4x4?)故 ???1 …..……….……..8/
四, 求表面积为a2而体积最大的长方体的体积.
解： 设长方体的长,宽,高为x,y,z，则问题变为求函数V?xyz,在条件
?(x,y,z)?2xy?2yz?2xz?a2下的最大值. …………………….…..2/
设F(x,y,z)?xyz??(2xy?2yz?2xz-a2), ………….….4’
??Fx(x,y,z)?yz?2?(y?z)?0由??Fy(x,y,z)?xz?2?(x?z)?06F得x?y?z?a ……………..6/
?z(x,y,z)?xy?2?(y?x)?06???(x,y,z)?2xy?2yz?2xz?a2只有一个驻点，故当x?y?z?666a时，最大值V?xyz?a3…………8/ 36
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