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大学工科数学分析期末考试_(试题)A

大学工科数学分析期末考试_(试题)A
大学工科数学分析期末考试_(试题)A

20XX年复习资料

专业:

班级:

科目老师:

日期:

一、填空题(每题4分,共20XX 分)

1. 设

ABC

L 是从

(1,0)

A 到

(0,1)

B -再到

(1,0)

C -连成的折线,则曲线积分

d d ||||

ABC L

x y

x y +=+? .

2. 设向量场222(1)(1)(1)A x x z i y x z j z x z k =++-+-,则向量场在点012

1M -(,,)处的旋度A =rot .

3. 若x y xe -=和sin y x =为某四阶常系数齐次线性微分方程的两个解,则该方程是

.

4. 函数(),(),(,)x x f x y ?ψ皆可微,设()(),()z f x y xy ?ψ=+,则

z z

x y

??-=?? . 5. 锥面

22

z x y +被圆柱面

222,(0)

x y ax a +=>截下的曲面的面积

为 .

二、单项选择题(每题4分,共20XXXX 分)

本题分数 20XX

得 分

本题分数 20XXXX 得 分

(多选不得分)

6.若

()()

0000,,,

x y x y f

f x

y

????都存在,则(,)f x y 在()00,x y ( )

(A )极限存在但不一定连续 (B )极限存在且连续

(C )沿任意方向的方向导数存在 (D )极限不一定存在,也不一定连续

7. 12,L L 是含原点的两条同向封闭曲线,若已知122

d d L y x x y

K x y -+=+?(常数),

则222d d L y x x y

I x y -+=

+?的值

( )

(A )一定等于 K (B )一定等于K -

(C ) 与2L 的形状有关 (D )因为

Q P

x y

??=??,所以0I =

8.∑为球面2222x y z a ++=外侧,Ω为球体2222x y z a ++≤,则有 ( )

(A )222254()3

d d x y z v a v a πΩ

Ω

++==??????

(B )2d x S ∑

??=

222

2

4114()3

3d d 3x y z S a S a π∑

++==???? (C )由于∑关于yOz 平面对称,所以d d 0x y z ∑

Λ=?? (D )20d d x y z ∑

Λ=??.

三、解答题(第20XXXX 题8分,其余每题20XXXX 分,共68分)

9. 求曲面积分d d d d x y z xz z x ∑Λ+Λ??,其中222: 1 (0)x y z z ∑++=≥,取上侧.

20XXXX. 已知()F x 在[]0,1上具有连续的二阶导数,()()()01, '11F F F ==,

计算()22d d D

I xyF x y x y ''=+??,其中22:1,0,0D x y x y +≤≥≥.

本题分数 68

得 分

20XXXX. 设函数()f t 具有二阶连续导数,函数22()u f x y =+满足

2222

221u u u u x y x y x x

???+-+=+???,求函数u 的表达式.

12.计算()2d d d d z x y z z x y ∑

+-??,其中∑是旋转抛物面()2

212

z x y =

+介于平面0z =及2z =之间的部分的下侧.

13. 求线性微分方程组

1

2

2

123

3

123

d

d

d

442

d

d

2

d

x

x

t

x

x x x

t

x

x x x

t

?

=

?

?

?

=-++

?

?

?

=--

?

?

满足

123

(0)(0)(0)1

x x x

===的

特解.

14. 设函数(,)x y θ在xoy 面上具有一阶连续偏导数,曲线积分

2d (,)d L

xy x x y y

θ+?与路径无关,且对任意

t 恒有

(,1)

(1,)(0,0)

(0,0)

2d (,)d 2d (,)d ,t t xy x x y y xy x x y y θθ+=+?

?

求(,)x y θ.

20XXXX. 设S 为椭球面的上半部分,点(),,P x y z S ∈,π为S 在点P 处的切平面,ρ为原点到平面π的距离,求()

d ,,S

z

S x y z ρ??

.

一、 填空题

1.2-; 2.234(,,)

; 3.(4)2220y y y y y ''''''++++=; 4.()'2()(),()'()y x f x y xy xy ?ψψ-+?; 5.3

π

二、 选择题

6. (D) 7.(A) 8.(D ) 三、

9 解:设220:0 (1)z x y ∑=+≤,取下侧,则

d d d d d d d d d d d d x y z xz z x x y z xz z x x y z xz z x ∑∑+∑

∑+=+-+??????.……… (3

分)

设222: 1 (0)x y z z Ω++≤≥,可得

02

d d d d 1d d d ()3

x y z xz z x x y z V π∑+∑

Ω+=?=Ω=?????.…………… (6分)

同时,0

d d d d 0x y z xz z x ∑

+=??,

…………………………………………………… (8分)

以原式

2

3

π=.…………………………………………………………………… (20XXXX 分)

20XXXX. 解:利用极坐标变换

()

()

()

11

12

2

3''23

22

20

1d cos sin "d cos sin d d "d 2I r F r r r r F r r r F r r π

π

θθθθθθ===

?????,………… (5分)

令2t r =,则

()()()()()()()11100011111

"d d ''1'd '11044444

I tF t t t F t F F t t F F F ??=

==-=?-+?=?????????.…………

(20XXXX 分)

20XXXX. 解:设22

t x y +,原方程化为2''()()u t u t t += (4)

解之得212cos sin 2u C t C t t =++-……………………………… (9分) 故222222122

u C x y C x y x y =+++-…………………………… (20XXXX

分)

20XXXX 解:在曲面∑上,有22

cos 1x y α=

++,22cos 1x y γ

++,因此 ……(2分)

(

)()(22222

22d d d d cos cos d d 11z x y z z x y z x z S z x

S

x y x y αγ∑

∑∑?

?

???+-=+-=+???++++?

??????……(5

分)

()

()()

2

22

221

1d d .

4

2xy D x y x x x y x y ????

=-++?--+????????

?? ……(7分)

()

2222222

20011d d d cos d 822xy D x x y x y r r r r πθθπ????=

++=+= ????

??????? ……(20XXXX

分)

20XXXX. 解 解得特征根为210321===λλλ,, ……………… (2分)

对应的特征向量是,021,122,201321???

??

??=????? ??=????? ??=r r r ……………… (5分)

所以通解为:()X t =????

?

?

?022202122t t

t

t t e e e e e ???

?

?

??321c c c . ………………… (7分) 代入初值条件得1231/2

01/2

c c c =??

=??=?,特解为:21223

()1/2/2()()1t t x t e x t e x t ?=+?=??=?.……… (20XXXX

分)

20XXXX.

因为积分与路径无关,所以

2,x x

θ

?=?即

2(,)()x y x c y θ=+. …………………………(3

分)

又 (,1)

1

1

22(0,0)002d (,)d [()]d ()d t xy x x y y t c y y t c y y θ+=+=+???

(1,)

(0,0)

2d (,)d [1()]d ()d t t t

xy x x y y c y y t c y y θ+=+=+???

则 1200()d ()d t

t c y y t c y y +=+??, …………………………(7分)

两边同时求导可得21()()21t c t c y y =+?=-.

因此 2(,)21x y x y θ=+- .…………………………(20XXXX 分)

20XXXX.解: 设(),,x y z 为π上任意一点,则π的方程12

2

xX

yY

zZ +

+=, 于是()1

22

2

2222,,44Ax By Cz D x y x y z z A B C ρ-

+++??=++ ?

++??

.……………………………………………(3

分)

由曲面方程知2212

2x y z ??

-+ ?

??

222122z

x

x y ????-+ ???,22212

2z y x y ?=

???

-+ ?

??2

2222

2

4d 1d 212

2x y z z S x y x y σσ

--??

????=++ ? ?????????

-+ ?

??,

………………………………………(5分)

故()())

2222222200113

d d 4d d 4d ,,44442S

S

D z x y S z S x y r r r x y z πσθπρ=++=--=-=??

?????………(8分)

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