2010年全国大学生数学专业及高等数学竞赛试题及解答(3)

2019-04-14 10:56

(2)计算I????axdydz??z?a?dxdyx?y?z222222,

其中?为下半球z??a?x?y的上侧,a?0.

2解法一. 先以x?y?z2?22122??a代入被积函数,

?z?a2axdydz??z?a?dxdy1z? ???axdyd?I???a?a?d,x dy222?x?y?a?补一块有向平面S:?,其法向量与z轴正向相反,

?z?0利用高斯公式,从而得到

?1?22I????axdydz??z?a?dxdy???axdydz??z?a?dxdy?

a??S???+S-??1?2???????a?2?z?a??dxdydz???adxdy?, ??a??D?其中?为?+S围成的空间区域,D为z?0上的平面区域x?y?a,

?2221?2322?于是I???3a??a?2???zdxdydz?a?a?

a?3??2?a014 ???a?2?d??dr?00?a?a2?r2zdz

? ???2a3.

解法二. 直接分块积分

I1?1222axdydz??2a?x?ydydz, ??????a?Dyz222其中Dyz为yOz平面上的半圆y?z?a,z?0. 利用极坐标,得

11

I1??2?d???2?a02a2?r2rdr???a3,

312I2????z?a?dxdy

a?21?222? ???a?a??x?y?dxdy,

???aDxy?其中Dxy为xOy平面上的圆域,x?y?a, 用极坐标,得

a12?I2??d??2a2?2aa2?r2?r2rdr

0a0222?? ??6a3,

因此I?I1?I2???2a3.

(3)现要设计一个容积为V的圆柱体的容积,已知上下两低的材料费为单位面积

a元,而侧面的材料费为单位面积b元.试给出最节省的设计方案:即高与上下底

面的直径之比为何值时,所需费用最少?

解:设圆柱体的高为h,底面直径为d,费用为f, 根据题意,可知??24V?d?2h?V, dh????2?2?d?f?a?2?????b??dh

?2? ????12?ad?bd?h ?2?11?1????ad2?bdh?bdh?

22?2?1??33ad2?bdh?bdh 2

12

3?32322?ab??dh? 23?323?4V??ab???, 2???当且仅当ad?bdh时,等号成立,

22ha?, db故当

ha?时,所需要的费用最少. db1?11?(4)已知f?x?在?,?内满足f??x??求f?x?. 33sinx?cosx?42?解:f??x??1?sin3x?cos3xdx

2?11sinx?cosx?? ????dx, 223?sinx?cosx2sinx?sinxcosx?cosx?111?sinx?cosxdx?2????dx

sin?x??4??x?1 ?lntan4?C1,

22sinx?cosxsinx?cosxdx?dx ?sin2x?sinxcosx?cos2x?112?sinx?cosx??22??sinx?cosx??1d?sinx?cosx? ?2?2?sinx?cosx??1?2?sinx?cosx2dx

?2arctan?sinx?cosx??C2

13

21所以,f?x??lntan32二、

求下列极限.

x??4?2arctan?sinx?cosx??C. 23??1?n? (1)limn??1???e?;

?n?????n???a?b?c? (2)limn???3??1n1n1n??,其中a?0,b?0,c?0. ???n??1?n???1?x?解:(1)limn??1???e??limx??1???e?

?x??????n???x???n???? ?limx???e?1?xln1????x??e1xx

1??1????11?ln?1?x???????x???x???1?x ?limx???1?2x1????x??

1?1?ln?1???x?1?x ?elim?

x???1?2x111??1?xx?1?x?2 ?elim

x???123x1?21?x?e? ?lim

x???12x2

14

e1e ??lim??. 2x???22?1??1??x???a?b?c?(2) limn???3??1n1n1n????lim?a?b?c?x????3????lnx???n1x1x1x?? ???x111xxxa?b?c ?limex???xln111xxxa?b?clim3 ?e31x,

lnx???lima?b?c3 1x11x1x1x11?1??1?xxx111?alna?blnb?clnc???2??x??xxx??lima?b?c x???1?2x11?1?xxx?lim1alna?blnb?clnc? 11?x????ax?bx?cx?1?1?lna?lnb?lnc??ln3abc, 31n1n1n??a?b?c故limn???3??m???3abc. ???1nkn???a一般地,有lim?k?1n???m??

????ma1a2?am,其中ak?0,k?1,2,?,m, ???15

n


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