第七章 重积分作业参考解答

2020-05-24 10:07

第七章 重积分作业参考解答

习题7.1 4.(1)

解答:在积分区域D内,

?1?x?y?2,?ln(x?y)?1,[ln(x?y)]2?ln(x?y),

5.(1) 解答:?2>ln(x?y)d?[ln(x?y)]d? ????DD?4?x2?y2?3?, 43?2?2?2222?? ??sin(x?y)?1,圆环积分区域的面积为S?442222?222故 ????sin(x?y)d??42D5.(3)

解答:?0?x?4,0?y?8,

?ln4?ln(4?x?y)?ln16,积分区域的面积为S?4?8?32

故 32?111 ???d??32?ln16Dln(4?x?y)ln4即

习题7.2

8116 ???d??ln2Dln(4?x?y)ln2x22y1.(2)解答:原式=?dy?xedx=?edy?xdx=?e[]?ydy

11?y1?y232yy33y32y3y33311333y2dy??eydy3?[ey]1 ??e

112223y3?1.(3)解答:原式=

127(e?e) 2?dy?011yx3?1dx

变化积分次序为:原式= =

?dx?01x20x3?1dy

?10x3?1?x2dx

1133=?x?1d(x?1) 301231=?[(x?1)2]0 33=

342?2 9xy22.(2)解答:积分区域如图1,

原式=

?dx?xye0011dy

1.4111xy22=?dx?ed(xy) 20011x =?(e?1)dx

201x1 =(e?x)0

21 =(e?2)

22.(4)解答:积分区域如图2:

交点为(1,1),(-1,-1),积分区域看做Y型:

1.51.00.51.21.00.80.60.40.20.00.00.20.40.60.81.01.21.4图 1 ?1?y?1,y2?x?3?2y2

原式=

?11?1dy?2y3?2y2(y?y)dx 0.51.01.520.51.01.52.02.53.03.5图 2

= =

??11?2y(y2?y)[x]3dy y22??1(y2?y)(3?3y2)dy=?(3y2?3y4?3y?3y3)dy

?11=[y?33532341y?y?y]?1 524764= 52.(6)解答:积分区域如图3: 交点为(2,4),积分区域看做Y型:

540?y?4,原式=

y?x?6?y 232?40dy?y6?yyexdx

12001234567图 3

=

?40y[e]x6?yy2dy=?y[e046?y?e]dy=?(eye0y246?y?ye]dy=e6?9e2?4

y23.(1)解答:积分区域如图4, 原式=

3.(4)解答:积分区域如图5 原式=

图 4

1?10dy?f(x,y)dx

y1?10dx?x2f(x,y)dx??dx?x2f(x,y)dx

212x22

图 5

3.(5)解答:积分区域如图6 原式 =

图 6

1?1?y2?dy?010f(x,y)dx??dy?014?y21?1?y2f(x,y)dx??dy?124?y20f(x,y)dx

228.(2)解答:积分区域{(x,y)1?x?y?4}关于y轴对称,而sin(xy2)是x的奇函数,

y的偶函数,故

??sin(xy)d??0

D210.(1)解答:积分区域如图7

??f(x,y)dxdy??D2?0d??f(rcos?,rsin?)rdr

0a图 7

图 8

?10.(3)解答:积分区域如图8,

??f(x,y)dxdy??D20d???31cos??sin?01cos?0f(rcos?,rsin?)rdr

f(rcos?,rsin?)rdr

11.(2)解答:积分区域如图9,

?dx?013xxf(x,y)dy???d??4

图 9

图 10

?

11.(3)解答:积分区域如图10,

?10dx?1?x21?xf(x,y)dy?22?20d??11cos??sin?f(r2)rdr

12.(3)解答:积分区域如图11,

0.7?220dy?1?y2yyarctandx??04d??0arctantan??rdr

x1?0.60.50.40.30.20.10.00.00.20.40.60.81.01.2图 11

? =

?40d????rdr=??d??0401?10?2rdr=

641.41.213.(2)解答:积分区域如图12

1.00.81d??22??4?x?yD??401d???rdr

04?r220.60.40.20.00.00.51.01.52.0?21??141222??[ln(4?r)]ln2 dr = ?d??==004?r224820?1图 12

13.(3)解答:积分区域如图13,

??ln(1?x?y)d??D22?20d??ln(1?r2)rdr

0?11?2ln2?1122212122? = ?d??ln(1?r)dr=??{[(r?1)ln(1?r)]0?(1?r)0}=

022420图 13

1ex图 14

11x?x1x?x(e?e)?e??2 ==(e?e)dxdxdy0?0?e?x?0e14.(1)解答:积分区域如图14, 原式=

14.(3)解答:积分区域如图15,原式=

?0?1dy??1?y2y?1y2y301dx=?(?y?y)dy=(??)?1=

?162302

图 15

15.(2)解答:积分区域如图16 原式=

?1?1dx?2?x22?2?x((10?3x2?3y2)?4)dy

图 16

=

?2?0d??(6?3r2)rdr=?d??(6r?3r3)dr

000222?2=2??(3r?342r)0=6? 416.(2)解答:由题意知积分闭区域在极坐标系上变量范围为0????,1?r?1?sin?,故 原式=

??D1x2?y2d??

??0d??1?sin?1r21?rdr=?sin?d?=2

0?16.(3)解答:积分区域如图17,利用极坐标系知变量范围为0???

?2,0?r?1,


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