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2011年高等数学竞赛例题选讲

发布时间:2013-10-08 08:04:27  

高等数学竞赛例题选讲

一、一元微积分(A组考,B组考)

例1(8分)、设f(x)?lim(1?xn??13nn),试讨论该函数在 (??,??)内的可导性.

解:当x?1时,1?(1?x

1

n

n??3n--------------------------------------------------------------------1 )?2,1n1n注意到lim2?1,由夹逼定理得 当x?1时,f(x)?1;----------------------------------1 113同理,当x?1时,f(x)?xlim(1?)n?x.------------------------------------------1 n??x

显然,当x?1时,f(x)可导,---------------------------------------------------------------------1 33n

1?1x3?1?0,f?'(1)?lim?3,---------------------------2 而x?1时,因f?'(1)?limx?1?x?1x?1?x?1

所以,f(x)在x?1时不可导,同理,f(x)在x??1时也不可导.-----------------------2

x4f(x)?ln(1?x2)?x22?, 例2(8分)、设f(x)在x?0处具有二阶导数,且有lim6x?0x3

求f(0),f'(0),f''(0).

x4f(x)?ln(1?x2)?x22???,其中lim??0---------------------------------------------1 解:因x?0x63

ln(1?x2)?x2222?x??x,---------------------------------------------------------1 则f(x)??4x3

ln(1?x2)?x21?,-------------------------------------------2 于是,f(0)?limf(x)??lim4x?0x?0x2

1ln(1?x2)?x2?x4f(x)?f(0)?0-----------------------------------2 f'(0)?lim??lim5x?0x?0xx

f'(x)?f'(0)f(x)?f(0)2?2lim?.故f''(0)?lim---------------------------------------2 x?0x?0xx23

t1?x???1t4?4dt确定了y?y(x),求dy|t?2. 例3(8分)、?dx?y?eysin(t?2)?1?

解:t?2时,y?1 --------------------------------------------------------------------1 dx11dx1由 知 -----------------------------------------------------------------2 ???t?22dt8dt2t?4

由dydydy?eycos(t?2)?eysin(t?2) 知 ?e----------------------------------------3 dtdtdtt?2

∴dy??8e.--------------------------------------------------------------------------------2 dtt?2dx

dtt?2dydtt?2

例4(8分)、设F(x)为f(x)的原函数,当x?0时,f(x)F(x)?sin22x,且F(0)?1, F(x)?0,求f(x).

解: [F2(x)]??2F(x)F?(x)?2F(x)f(x)?2sin22x=1?cos4x ---------------------------2

1

F2(x)??2sin22xdx?x?sin4x?C ----------------------------------------------------- 2

4

1

F2(0)?C?1 ,故F2(x)?x?sin4x?1 -----------------------------------------------2

4

1?cos4x

.----------------------------------------------------------------- 2 ?f(x)?

1

x?sin4x?14xearctanx

dx. 例5(8分)、计算?2(1?x)

tt

解(法一):设x?tant,则原式=esintdt??edcost--------------------------------------3

??

ttttt

=?(ecost?ecostdt)=?ecost?esint?esintdt,-------------2

??

1t

e(sint?cost)?C.--------------------------------------------------------1 ?2

(x?1)earctanx

故 原式=?C.-------------------------------------------------------------------2

2

2?x

xearctanxearctanxxarctanx

??------------------------------4 解(二): 原式=? =222

?x?x(1?x)而esintdt?

t

=

xearctanx?x

2

??

1?x

2

de

arctanx

=

xearctanx?x

2

?

earctanx?x

2

??

xearctanx(1?x)

2

, -----2

移项整理得,原式=例6(8分)、已知

(x?1)earctanx2?x

2

?C.------------------------------------------------------------2

secx3

是函数f(x)的一个原函数, 求?xf?(x)dx. x

secxsecx(xtanx?1)

)??解 :由题意有f(x)?(---------------------------------2 xx2

3232?secx?原式?xf(x)?3?xf(x)dx?xf(x)?3?xd??----------------------------3

x??

secx

?x3f(x)?3(x2??2?secxdx)=xsecx(xtanx?4)?6lnsecx?tanx?C.------3

x

例7(8分)、设f?x?在?1,???内可导,f?0??1,其反函数为g?x?,且满足

?

x?f?x?

x

g?t?x?dt??2x?1?f?x?,(1)求?g?x?dx;(2)求f'?0?.

1

解:(1)将x?0,f?0??1代入(2)令t?x?u,则有

?

x?f?x?

x

g?t?x?dt??2x?1?f?x?,易得?g?x?dx?1----2

1

?

x?f?x?

x

g?t?x?dt??

f?x?

g?u?du??2x?1?f?x?-------------------3

'

上式两端关于x求导,注意到g[f?x?]?x,有xf?x???2x?1?f'?x??2f?x?-----------2

将f?0??1代入上式,解出f'?0???2.--------------------------------------------------------------1

例8(8分)、已知?1

?1[2(2x2?1)f(x)?f''(x)]exdx?2e,且f'(?1)?f'(1),

22求f(?1)?f(1). 解:?1

?12(2x2?1)f(x)exdx??exdf'(x)?2e-----------------------------------------------------2 ?1

2212?1

?12(2x2?1)f(x)exdx?[f'(x)ex]1f'(x)xexdx?2e-------------------------------2 ?1?2??1

12122?f(x)dxex?2?xexdf(x)?2e------------------------------------------------------------------2 ?112?1

?2e故f(?1)?f(1)?1.则2[f(x)xex]1------------------------------------------------------------2 ?1

111limf(x)limf(x)f(x)dx?1,求与?f(x)dx. 0x?001?x2x?0

1a?1 -------------------------2 解:令limf(x)?a,?

f(x)dx?b,则f(x)?20x?01?x

a?1]?a?b------------------------------------------------------1 ?1故有

a?li2?x?01?

x

1a?b??(?1)dx??a?b??1-------------------------------------------------------3 01?x24

18??, ?f(x)dx?b?1. ----------------------------------2 由上述两式解之得limf(x)?a?2例9(8分)、

若f(x)?x?0?0

例10(8分)、(1)证明:

(2)利用(1)计算

x???t??0?0xf(sinx)dx??2?0?f(sinx)dx ?

?xsin2008x. 20082008sinx?cosx??00证明:(1)左

移项得???0(??t)f(sin)tdt???f(sinx)dx??xf(sinx)dx-------------------3 ?f(sinx)dx--------------------------------------------------------------1 2?0

??sin2008x(2)原式??dx------------------------------------------------------------------1 20sin2008x?cos2008x

20082008???sinxsinx ?(?2dx?dx)-------------------------------------1 ?2008200820082008?02sinx?cosxx?cosx2sin??0xf(sinx)dx?

故原式?????.--------------------------------------------------------------------------------------------------1 4

?x例11(8分)、求曲线y?esinx,x?0与x轴所围成图形的面积A. 解:A?22sinx?sin2008x?cos2008x2008x??u2???20cos2008xdx------------------------------2 sin2008x?cos2008x???

??esinx??(?1)?xn?0?n???n??n(n?1)?e?xsinxdx---------------------------------------------3 ??

x?n??t?e

n?0??0esintdt ??esintdt(?e?n?)----------------------------------------3 ?t0n?0?t?1?1?e????01e??1esintdt? .-------------------------------------------------------------------2 ?2e?1?t

例12(10分)、某函数y?f(x)满足f(x)?0,f"(x)?0,f(0)?0,f(1)?2,其所示曲线与一条单调递增的直线y?kx在第一象限有唯一交点(t,f(t)),其中t?(0,2).

现记曲线y?f(x)与直线y?kx围成的平面区域绕x轴旋转一周形成的旋转体体积为V1,而它们与直线x?2围成的平面区域绕x轴旋转一周形成的旋转体体积为V2,试确定k,t之值,使V?V1?V2达到最小. 解:由题意知k?1313f(t)?0------------------------------------------------------1 t

1

V?V1?V2???[(kx)2?f2(x)]dx???[f2(x)?(kx)2]dx--------------------1 0tt23

??f(t)

t22(?x2dx??x2dx)???f2(x)dx???0t0

21t123t123tf2(x)dx 2f(t)??[tf2(t)?2]??f2(x)dx??f2(x)dx}--------------------------1 0t3t

41V'??(1?3)[tf'(t)?f(t)]f(t)-------------------------------------------2 3t

令g(t)?tf'(t)?f(t)----------------------------------------------------------------------------------1 t23由g'(t)?f"(t)?0知g(t)于[0,2)上单调递增,-----------------------------1 于是,t?0时,有g(t)?g(0)?0--------------------------------------------------------------1 令S'?0,得唯一驻点t?1-------------------------------------------------------------------------1 由极值的第一充分条件易知t?1,即k?2时,--------------1 V?V1?V2达到最小.

例13(8分)、就参数k,讨论曲线y?4lnx?k与y?4x?ln4x交点的个数.

tt4解:令lnx?t即x?e,考虑方程4e?t?4t?k?0 ①--------------------------1 13

令f?t??4e?t?4t?k,-------------------------------------------------1 t4

t3t2则f??t??4e?t?1 及 f???t??4e?3t?0②-------------------------1 ????

故f?t?有唯一驻点t?0,由②可知f?0??4?k是f?t?的最小值.-------------------1 又注意到limf?t????-----------------------------------------------------1 t??

当4?k时,方程①有两解,曲线有两个交点.----------------------------------1 当4?k时,方程①有一解,曲线有一个交点.----------------------------------1 当4?k时,方程①无解,曲线没有交点.--------------------------------------1

1?1有且仅有一个根,求k的取值范围. 2x

11解:设f(x)?kx?2?1,当k?0,f'(x)?k?3?0,f(x)为减函数-----------------2 xx

f(x)=??,limf(x)?0,故k?0满足题意-----------------------------------------2 又lim?例14(8分)、设当x?0时,方程kx?x?0x???

2,由f''(x)?0知其为极小点,------------2 k

23时有且仅有一个根, 若f(x0)?0,原方程或无解或有两解,仅当f(x0)?0,k?9

23.-------------------------------------------------------2 综上所述,k的取值范围为k?0及k?9当k?0,令f'(x)?0,得唯一驻点x0?

1n??

)对所有的正整数n都成立的最小的数?. n111n??

? -----------------------------------------------------2 解:由e?(1?) 解得??

11n

ln(1?)

nn

x

ln(1?x)?

11x?x令f(x)??,x?0,f'(x)?[ln(1?x)?] -----------2 22ln(1?x)xxln(1?x)?x

x

?x?1?

x?0, 记g(x)?ln(1?x)?,由x?0得g'(x)?3?x

(1?x)2

于是f'(x)?0从而x?0,f(x)单调减---------------------------------------------------------------3

1111

注意到limf(x)?,有x?0,f(x)?,则f()?,

x?022n2

11

而f()??,故?最小值为.---------------------------------------------------------------------------3

2n

12

例16(8分)、设x1?0,xn?1?(3xn?3),(n?1,2,3?),求limxn.

n??4xn

例15(10分)、求使得不等式e?(1?解: xn?1?又

122(xn?xn?xn?3)?xn?xn?xn??2则数列有下界,-----------------4 4xnxn

xn?112

故数列单调减少,易得limxn?2.-----------------------4 ?(3?4)?1,

n??xn4xn

1

?xn?1?xn例17(10分)、设数列{xn}、{yn}满足0?x1??,xn?1?sinxn,yn???

?xn?

(n?1,2,3,?),请问{xn}、{yn}收敛吗?若收敛,求limxn,limyn;若发散,说明理由.

n??

n??

答: ?x2?sinx1,?0?x2?1,则n?2,0?xn?1?sinxn?xn,{xn}单减有下界------2 根据单调有界定理知{xn}收敛,--------------------------------------------------1

0----------------2 令limxn?A,在xn?1?sinxn两边取极限得A?sinA,有limxn?

n??

n??

?sint?t2

先考虑 lim???et?0

?t?

故limyn?e

n??

?16

1

sint

limlnt?0tt

1

?e

sint

?1

limtt?0t

?et?0

lim

sint?tt

?et?0

lim

cost?13t

?e--------------3

?

16

,从而{yn}收敛.-------------------------------------------------2

例18(8分)、f(x)?arctanx在[0,b]上由拉格朗日中值定理得中值?,求lim

b?0

?2

b2

.

解:由拉格朗日中值定理得arctanb?12,即?2?b?arctanb------------------3

arctanbb1??故lim

b?0

?

2

b2

?lim

b?arctanbb?arctanb

?lim?limb?0b?0b2arctanbb?0b3

1?

1

2?1.------------------5

33b2

二、多元微分学(A组考,B组考)

?(x2?y2)xy,(x,y)?(0,0)例1(8分)、证明:f(x,y)??在(0,0)连续. 0,(x,y)?(0,0)?

证明:令limf(x,y)?A,则A?ex?0y?0x?0,y?0limxyln(x2?y2)------------------------------------2 12(x?y2)|ln(x2?y2)|-----------------------------------2 2

122222222?lim(x?y)|ln(x?y)|?0----2 tlnt?0因lim(x?y)ln(x?y)?lim,x?0,y?02x?0t?0?

y?0而0?|xyln(x?y)|?22

由夹逼定理知:A?e?1?f(0,0),原题得证. -----------------------------------2 例2(8分)、设?(x,y)连续,?(x,y)?x?y?(x,y),讨论?(x,y)在(0,0)处的可微性. 0

x?(x,0),且lim?(x,0)??(0,0)--------------------------------------------2 x?0x?0x

x若?(0,0)?0,因lim不存在,故?x(0,0)不存在,从而?(x,y)在(0,0)处不可微-----1 x?0x

若?(0,0)?0,则?x(0,0)?0,同理?y(0,0)?0--------------------------------------------------1

解:?x(0,0)?lim因0?

故lim?

?lim(x,y)??lim(x,y)??0-----------------------------2

(x,y)?(x,y)?0,即?(x,y)在(0,0)处可微.--------------2

,试确定常数 ,使

. 例3(8分)、设

解:,

-------------------------------2

-------------------------------3 由

,可得

.----------3

例4(8分)、设u(x,y)二阶偏导数连续,且uxx?uyy?0, u(x, 2x)?x,ux(x, 2x)?x2, 求uxx(x, 2x),uxy(x, 2x),uyy(x, 2x)(ux表示u对x的一阶偏导数,其他类推).

1

2

这两个等式,对x求导得uxx(x,2x)?2uxy(x,2x)?2x, uyx(x,2x)?2uyy(x,2x)??x.-------------2 ?uy(x,2x)?(1?x2)--------3 解:等式u(x,2x)?x两端对x求导,得ux(x,2x)?2uy(x,2x)?1,由已知条件得uxx?uyy,uxy?uyx, 故解得uxx?uyy??45x, uxy?x . ---------3 33

例5(8分)、设z?

解:z??10xy?tf(t)dt,0?x,y?1,若f(t)为连续函数,求zxx?zyy. 1

xy?xy0(xy?t)f(t)dt??(xy?t)f(t)dt--------------------------------------2

xy

xy

0?xy[?zx?y[?f(t)dt??f(t)dt]??tf(t)dt??tf(t)dt------------------------1 xyxy011xy

f(t)dt??f(t)dt] ---------------------------------------------2 xy1

zxx?2y2f(xy),由对称性知zyy?2x2f(yx)---------------------------------2 故zxx?zyy?2(x2?y2)f(xy). ---------------------------------------------1 例6(10分)、已知f?u?具有二阶导数,且f?(0)?1, y?y(x)由y?xey?1?1所确定,

dzd2z设z?f(lny?sinx),求x?0,x?0. dxdx2

解:在y?xey?1?1中, 令x?0 得y(0)?1 . ------------------------------------1 而由y?xey?1?1两边对x求导得 y??ey?1?xey?1y??0---------------------------1 再对x求导得 y???ey?1y??ey?1y??xey?1y?2?xey?1y???0-------------------------1 将x?0,y?1代入上面两式得 y?(0)?1,y??(0)?2. ------------------------------2 dzy??(?cosx)f?(lny?sinx),-----------------------------------------------1 dxy

d2zy?y??y?y?2

2 2?(?cosx)f??(lny?sinx)?(?sinx)f?(lny?sinx)--------------2 dxyy2

dzd2zy??(0)?2,f?(0)?1代入上面两式得将y(0)?1,y?(0)?1,x?0?0,x?0?1.---2 dxdx2

?2z?2z1例7(10分)设f?u?在?0,??

?内二阶可导,z?f, 满足2?2?22?x?yx?y

若f?1??0,f??1??1,求f?u?的函数解析式. ?zx?zy?f??u?,?f??u?-------------------------------------------------------------------------1 ?xu?yu

x2

u??2zx2x2y2?f???u?2?f??u?2?f???u?2?f??u?3------------------------------------------2 2?xuuuu

?2zy2x2

同理2?f???u?2?f??u?3-------------------------------------------------------------------------1 ?yuu

1?2z?2z1代入2?2?2得uf??(u)?f?(u)?,即[uf'(u)]'?(lnu)'----------------------3 u?x?yu

lnuc11?f?(u)??,由f?(1)?1,得c1?1,于是f(u)?ln2u?lnu?c2,-------------------1 uu2

1由f(1)?0,得c2?0故f(u)?ln2u?lnu.---------------------------------------------------------1 2

例8(10分)、已知函数z?f(x,y) 的全微分dz?2xdx?2ydy,并且f(1,1)?2.求

y2

f(x,y)在椭圆域D?{(x,y)x??1}上的最大值和最小值. 42

解:于是 z?f(x,y)?x2?y2?C,-------------2 dz?2xdx?2ydy?d(x2?y2?C),

再由f(1,1)?2,得 C=2, 故 f(x,y)?x2?y2?2.--------------------------------1 令fx?2x?0,fy??2y?0得可能极值点为(0,0),且f(0,0)?2-------------------1 y2

?1上的情形: 再考虑其在边界曲线x?42

y2

?1),-----------------------------1 令拉格朗日函数为L(x,y)?f(x,y)??(x?4

??Lx?2(1??)x?0,?1?解 ?Ly?(?2??)y?0, --------------------------------------------------1 2??2y2

?1?0?x??4

得可能极值点为(0,?2),(?1,0) ,且f(0,?2)??2, f(?1,0)?3,-----------------2 2

y2

可见z?f(x,y)在区域D?{(x,y)x?最小值为-2.---------2 ?1}内的最大值为3,4

例9(10分)、当x?0,y?0,z?0时, 求函数u?lnx?2lny?3lnz在条件x2?y2?z2?6r2上的最大值, 并证明对任意的正实数a,b,c成立不等式2

?a?b?c?ab2c3?108??. 6??

解: 令F(x,y,z)?lnx?2lny?3lnz??(x2?y2?z2?6r2)--------------------1 1?F?(1)?xx?2?x?0

??F?2?2?y?0(2)?yy有?------------------------------------2 ?3(3)?Fz??2?z?0z?

2222?(4)?x?y?z?6r?0

由(1),(2),(3), 得y2?2x2,

代入(4),得 x?r,

可知最大值为u(r,6z2?3x2------------------------------------------2 y?2r,z?r及P(r,2r,r)-------------------------1 ?lnr?2ln(2r)?3ln(3r)?ln(6r6)------------------1

62r,3r)222

2462?x?y?z6即 lnx?2lny?3lnz?ln(6r),亦即 xyz?(6)??6????-------2 ?

?a?b?c?222令x?a,y?b,z?c, 于是ab2c3?108??.--------------------------2 6??6

三、空间解析几何(A组考,B组不考)

?b. 例1(8分)、设非零向量a,b,求证:lim(|a?tb|?|a|)?prja??

解:左

?

t?0??21?t?0tt?0?????右.--------8

?y?z?1?0 ?x?2z?0例2(8分)、在已知平面?:x?y?z?1?0内,求一直线l通过已知直线L:?

与已知平面?的交点且垂直于已知直线L.

?x?y?z?1?0?解:联立方程组?y?z?1?0,易得L与?之交点P(0,?1,0)----------------------------------2 ?x?2z?0?

L的方向向量为s?(0,1,1)?(1,0,2)?(2,1,?1),------------------------------------------------------2 可求得过P点且与已知直线L垂直的平面?方程为2x?y?z?1?0.-----------------------2

?x?y?z?1?0由题意知,所求直线l应为平面?与平面?的交线,其方程为?. ---------2 2x?y?z?1?0?

?2x2?y2?z?4例3(8分)、(1)求空间曲线?:?2 在xoy面的投影曲线L的方程; 2?x?y?z?0

(2)求以L为准线,母线与向量s=(1,0,?1)平行的柱面方程.

?2x2?y2?z?422z,解:(1)对?2,消得投影柱面方程x?2y?4,-------------------1 2?x?y?z?0

?x2?2y2?4故投影曲线L的方程为? -----------------------------------------------------------2 z?0?

(2) 在所求柱面取M(x,y,z),由题意必有M0(x0,y0,0)?L,使得M0//s---------2 ?x02?2y02?4?22有?x?x0y?y0z 化简得柱面方程(x?z)?2y?4.------------------------3 ???0?1?1

例4(8分)求以直线x?y?z为对称轴,半径R?1的圆柱面方程.

解:在圆柱面上任取一点M(x,y,z),过点M(x,y,z)且垂直于轴的平面为 (X?x)?(Y?y)?(Z?z)?0--------------------------------------------------------------------------2 轴方程的参数式为X?t,Y?t,Z?t代入平面方程得t?

故该平面和轴的交点为M1(x?y?z ------------------------1 3x?y?zx?y?zx?y?z,,)--------------------------------------1 333

222则M0M1的长等于半径R?1------------------------------------------------------------------------------1 故由距离公式得(2x?y?z)?(?x?2y?z)?(?x?y?2z)?9.----------------3

y?bz?c,)?0的所有切平面都通过定点. x?ax?a

''''FuFv(b?y)Fu?(c?z)Fv'''证明:由题意知,Fx?,Fy?,Fz?, 2x?ax?a(x?a)

故曲面过任一点(x,y,z)切平面的法线向量可选为例5(8分)设F(u,v)可微,求证:曲面F(

n?{(b?y)Fu?(c?z)Fv,(x?a)Fu,(x?a)Fv} …(4')

注意到向量{a?x,b?y,c?z}? …(2)

而(x,y,z)在曲面上,故(a,b,c)也在曲面上,原题得证. …(2)

例6(8分)求证:在曲线x?t4,y?t2,x?t的切线中,与平面x?y?z?1平行的切线有且仅有一条.

证明:曲线的切向量为(4t3,2t,1),若其与(1,1,1)垂直,则4t?2t?1?0----------------2 令f(t)?4t3?2t?1,则f'(t)?12t2?2?0,且limf(t)???,limf(t)???,-----3 t???t???''''''3

知4t?2t?1?0在(??,??)内有且仅有一个非零根------------------------------1 又t(4t3?2t?1)?(t4?t2?t)?0,则此根不满足x?y?z?0,故原命题成立.------2 例7(8分)若点M0(x0,y0,z0)是光滑曲面F(x,y,z)?0上与原点距离最近的点,试证:过点M0的法线必定通过坐标原点. 3

?minf(x,y,z)?x2?y2?z2

证明:考察条件极值问题?-----------------------------1 F(x,y,z)?0?

构造辅助函数L(x,y,z,?)?f(x,y,z)??F(x,y,z),-------------------------------1 按题意f(x,y,z)在点M(x0,y0,z0)达到条件极小值,必满足

Lx(M0)?2x0??Fx(M0)?0,Ly(M0)?2x0??Fy(M0)?0,Lz(M0)?2x0??Fz(M0)?0于是向量(x0,y0,z0)与曲面F(x,y,z)?0在点M0处的法向量(Fy,Fy,Fy)0平行,-------5 故曲面F(x,y,z)?0在点M0处的法线通过通过原点.------------------------------1

222例8(8分)求函数f(x,y,z)?40?x?2y?3z在点M0(?3,3,?2)处沿的方向导数,其中为f(x,y,z)?1过M0处的内法向量.

2?2222解: gradf(M0)?(fx,fy,fz)0??[(40?x?2y?3z)3(x,2y,3z)]M0?(2,?4,4)-3 3

令F(x,y,z)?39?x2?2y2?3z2 ,则可取(2,?4,4),----------------------------2

????f?故 M0?gradf(M0)?en?gradf(M0)?6.-----------------------------------3 ?n

23例9(8分)设x?y?z?3确定了隐函数z?z(x,y),求该隐函数在点(1,1)处方向导数

的最大值M.

解:当(x,y)?(1,1)时,z?1------------------------------------------------------------------------------1

设F(x,y,z)?x?y?z?3-----------------------------------------------1 则Fx?1,

故M?

2312Fy?2x,Fz?3z2 ,于是zx(1,1)??,zy(1,1)??----------------3

33?-------------------------------------------3

四、多元积分学(A组考,B组仅考二重积分)

例1(7分)计算I?

?

asin?

e

?y2

dyb2?y2a2?y

e2

?x2

dx??

bsin?

asin?

e

?y2

dy?

b2?y2

ytan?

e?xdx,其中

2

,且a,b,?均为常数.

2

解:积分区域D为a2?x2?y2?b2与0?y?xtan?的公共部分。取极坐标计算,有

0?a?b,0???

?

I??

asin?

e

?y2

dy?x2

dx??

2

bsin?

asin?

e

?y2

dyytan?b

2

?x2

dx

xtan?

???e?(x

D

2

?y2)

dxdy???e?rrdrd???d??re?rdr

D

a

2

2

?

??

?

?a?e1?r2?e?b2

d???ed(?r)??d?

a022

b

2

2

e?a?e?b

??.

2

?x,y?0?x?1,0?y?1?,求??f?x,y?y?x2dσ.例2(8分)设f?x,y??Max?x,y?,D??

D

D1???x,y0?x?1,x?y?1?

2

D

解:将区域D分成三块:D2??x,y?0?x?1,x?y?x--------------------------------2

D3

则原式?

1

1

2

D1

D2

???x,y?0?x?1,0?y?x???

2

2

2

D3

??y?y?x?d????x?y?x?d????x?x

2

2

1

x

2

1

x

?y?d?------------------------2

x2

??dx??y?yx?dy??xdx?2?y?x?dy??xdx?

x

1

?x

2

?y?dy----------------2

351?x1x?1x2x3x4?x5?114

???????dx????x??dx??dx ? .-------------2 030022322240????

y2?2

?,x?y?1且x?0,

例3(8分)设f?x,y??? x

?其它.?0,

??f?x,y?dxdy,其中D???x,yx

D

2

?y2?2y.

?

解:记区域D1??x,y?x?y?2y,x?0,------------------------------------------2

2

2

??

??

D

π

2sin?y

f?x,y?dxdy???arctandxdy??π2?d??rdr----------------------------------2

1x6D1

π

1?1?2

????4sin??1?d??π2????cos2??d?--------------------------------------2

2?6?2

?

π

?23?121?212

?????sin2???πsin2?d???.-------------------------2

2188?4?

?26

6

π2π6

例4(10分)若f(x,y)在D1:y?x?2?y,0?y?1;D2:0?x?1,x?y?2?x内连续,

(1) 证明:

??f(x,y)d????f(y,x)d?,??f(x,y)d????f(y,x)d?;

D1

D2

D2

D1

(2)

利用上式,求I?(1)证明:

D1?D2

??

2?yy

?.

2?xx

??

D1

f(x,y)d???dy?

D2

1

f(x,y)dx??dx?

D1

1

f(y,x)dy???f(y,x)d?--3

D2

同理,

??f(x,y)dxdy???f(y,x)dxdy------------------------------1

D1

(2)由(1

)知I? ?

D2

??D2

d??

??

D1

?

故I?

D1?D2

??

?---------------------------2

1

d?-------------2 2D1???D211

?(a?b).---------------------------------2 (a?b)dxdy??42D1?D2

n??

x2?y2?1

?

例5(7分)设f (r)在[0,1]上连续,则lim证明:

???x

2

?y2

?

n

f

x

2

?y2dxdy?0.

1

?

x2?y2?1

???x

0?r?1

2

?y2

?

n

f

x

2

-------2 ?y2dxdy??d??r2n?1f?r?dr?2??r2n?1f?r?dr,

?

2π1

设M?maxf?r?,则0?

2

x?y?1

??

?

x2?y2?f

n

2

dxdy?2?M?r2n?1dr?

1

2?M

2n?2

注意到:lim

1

?0,于是由夹逼定理可知要证结论成立.-------------------------5

n??n?1

例6(8分)若连续函数f(x,y

)满足f(x,y)?

求解:令

y2009

1?x2?y2?2x

??

f(x,y)d??1,

1?x2?y2?2x

??

f(x,y)d?.

f(x,y)d??A-------------------------------------------------------------------1

?

1?x2?y2?2x

??

则A?

1?x2?y2?2x

??

1)d??A

1?x2?y2?2x

??

y

2009

d??2?3d??

2cos?

1

(r?1)rdr-------3

?

16??3

??3cos?d??4?3cos2?d??------------------------------------------------------1

0309

??

?16?.----------------------3 ??3(1?sin2?)dsin??2?

3(1?cos2?)d??03309

例7(10分)设f(x,y)在单位圆域有连续的偏导数,且在边界上取值为零,而f(0,0)?1,

?f?f?y?x?y若D为圆环域:?2?x2?y2?1,求lim dxdy. 22????0?x?yDx

解:令x?rcos?,y?rcos?, ?f?f?x?f?y?f?f?f?f?r(?)?r(cos??sin?)?x?y则r-------------------------------2 ?x?y?r?x?r?y?r?x?y

f(x,y)在单位圆的边界上取值为零,则f(cos?,sin?)?0-------------------------1 I???Dxfx?yfyx2?y2

1rdxdy???D?frdrd? --------------------------------------------1 r2??d??02?2??fdr??f(rcos?,rsin?)|1?d?--------------------------------------2 ??r0

2???2?

0f(cos?,sin?)d???0f(?cos?,?sin?)d?----------------------------------1

?0?2?f(?cos?*,?sin?*),?*?[0,2?],-------------------------------------2 故limI??2?f(0,0)??2?.--------------------------------------------------1 ?

例8(8分)设函数f (u)连续,在点u = 0处可导,且f(0)= 0,f??0???3, 求limt?0??01πt4

x2?y2?z2?t2???f1

πt4

?x2?y2?z2dxdydz. ?解:记G?t??x2?y2?z2?t2???fx2?y2?z2dxdydz,应用球坐标,并同时注意到积分区域t?与被积函数的对称性,有

8G?t??4πt?2

0d??sin?d??f?r?r2dr?200?t4?f?r?r2dr0t4-------------------------------------4

4f?t?t2f?t??f?0??lim?lim?f??0???3.于是有limG?t??lim-------443t?0t?0t?0t?0tt4t

例9(8分)设f(x)在(??,??)内具有一阶连续偏导数,L是上半平面(y?0)内的有向分段光滑曲线,起点为(a,b), 终点为(c,d), 1x22记I??[1?yf(xy)]dx?2[yf(xy)?1]dy, Lyy

(1)证明曲线积分I与路径无关;(2)当ab?cd时,求I的值.

1x22解:(1)记P(x,y)?[1?yf(xy)],Q(x,y)?2[yf(xy)?1],则 yy

Qx?f(xy)?xyf?(xy)?y?2?Py, 于是P(x,y),Q(x,y)满足:在y?0时,04?f?r?r2drtQx?C,Py?C且Qx?Py ,所以曲线积分I与路径L无关--------------------------4

(2)由于曲线积分与路径无关,取L为从(a,b)到(c,d)的折线段,于是

c1d(c,b)(c,d)cI??P(x,b)dx??Q(c,y)dy??(?bf(xb))dx??(cf(xy)?2)dy-------2 abb(a,b)(c,b)y

cbcdcdc?acccaca???f(t)dt??f(t)dt????f(t)dt????.--------------2 abbcabbdbdbdb

?3x2?y2?z?6例10(8分)(1)求空间曲线?:?2 在xoy面的投影曲线L的方程; 2?x?2y?z?0

2ydx?xdy(2)若取上述L为顺时针方向,求?. L2x2?3y2

?3x2?y2?z?622z,解:(1)对?2,消得投影柱面方程2x?3y?6----------------------2 2?x?2y?z?0

?2x2?3y2?6故投影曲线L的方程为? --------------------------------------------2 z?0?

1(2)原式??2ydx?xdy------------------------------------------------------------------1 6L

Green1 .----------------------------------------3 ???

3d???62x2?3y2?6

ydx?xdy22L,其中圆周??x?1?y?2,L的方向为逆时针方向. L2x2?y2

解:由于x?y?0时,被积函数无意义,故L所包围的区域不满足格林公式的条件,作一小圆挖去原点?0,0?,作逆时针方向的圆周l:x?rcos?,y?rsin?,0???2? 使l全部被L所包围,在L和l为边界的区域D内,根据格林公式,有

??Q?P?ydx?xdyydx?xdy?? ?dxdy??????x?y?L2x2?y2l2x2?y2?D1?例11(8分)求-------------------------3?Px2?y2?Q∵ ,故上式为零----------------------------------3 ??222?y?xx?y

22222??rsin??rcos?ydx?xdy12??d???d????.-----2 ∴原式??2?l2x2?y2?0?02r2例12(10分)设函数Q(x,y)在x O y平面上具有连续一阶偏导数,曲线积分?2xydx?Q(x,y)dy与路径无关,并且对任意的t恒有

?2xydx?Q(x,y)dy??2xydx?Q(x,y)dy,求Q(x,y). L(t,1)(1,t)

(0,0)(0,0)

解:由曲线积分与路径无关知 ?Q??(2xy)?2x,-------------------------------------------------------------------------2 ?x?y所以Q(x,y)?x2?C(y),其中C(y)为待定函数。 又?(t,1)(0,0)2xydx?Q(x,y)dy??t2?C(y)dy?t2??C(y)dy--------------------------2 00tt001??1?(1,t)(0,0)2xydx?Q(x,y)dy???1?C(y)?dy?t??C(y)dy--------------------------------2 1t根据题设,有 t2??C(y)dy?t??C(y)dy,--------------------------------------------------------------1 00

上式两边对t求导,得到

2t?1?C(t),于是知C(t)?2t?1,即C(y)?2y?1,------------------------------2

故Q(x,y)?x?2y?1.--------------------------------------------------1 2

例13(12分)设?'(y)连续,在围绕原点的任意分段光滑简单闭曲线L上, ?(y)dx?2xydy

2x2?y4L恒为常数,

(1)对右半平面x?0内的任意分段光滑简单闭曲线C,有???(y)dx?2xydy

2x?y24C?0;

(2)求函数?(y)的表达式.

证明:(1)如图,将C分解为:C?l1?l2,另作一条曲线3围绕原点且与C相接,

???l2?l32x2?y4?0.-----------4 2x2?y4

?(y)2xy,Q?解:(2) 设P?,P,Q在单连通区域x?0内具有一阶连续偏导数, 24242x?y2x?y

?(y)dx?2xydy由(1)知, ?在该区域内与路径无关,故当x?0时,总有Qx?Py----2 L2x2?y4

?4x2y?2y52x2??(y)???(y)y4?4?(y)y3

由Qx?, Py?,------------------------2 242242(2x?y)(2x?y)

???(y)??2y,① 得? ----------------------------------------2 435??(y)y?4?(y)y?2y.  ?② 22由①得?(y)??y?c,将?(y)代入②得 c?0,从而?(y)??y.------------------2 则?(y)dx?2xydy2x2?y4C??(y)dx?2xydyl1?l3??(y)dx?2xydy例14(10分)设C是取正向的圆周(x?1)?(y?1)?1,f (x)是正的连续函数, 证明:22Cxf(y)dy?ydx?2?. f(x)

证明:由格林公式有Cxf(y)dy?

其中D是由 ( x – 1 )2 + ( y – 1 )2 = 1所围成的区域。而 ?y1?dx????f(y)?dxdy,-------------------------2 ?f(x)f(x)?D?

2??Df(x)dxdy??dx?021??(x?1)21??(x?1)2f(x)dy?2?f(x)?(x?1)2dx,----------------------------2 0

??f(y)dxdy??dy?D021?1?(y?1)221??(y?1)f(y)dx?2?f(y)?(y?1)2dy,--------------------------2 02

则左???f(x)dxdy???f(y)dxdy,----------------------------2 DD??1?1?f(y)?d??f(x)?d????2d??2?.-----------------2 ????????f(x)?f(x)?D?D?D

x2y2

??z2?1的上半部分,点P(x,y,z)??,?为?在P例15(8分)设?为椭球面22

zdS点处的切平面,?(x,y,z)为点O(0,0,0)到平面的距离,求??. ?(x,y,z)?

1?xXyY

x2y2

2?z)2???zZ?1,?(x,y,z)?(?解: ?的方程为4422x2y2

由z?

?(?),?22

则原式??

112?3222(4?x?y)d??d??r)rdr??. ----------------4 ???004x2?y2?242

tx2?y2

例16(10分)设?为夹在两个平面z?0和z?(t?0)之间的抛物面z?, 22

1?x2?y2

求证:??dS?(2ln2?1)?.

3

?(1?x2?y2)2

?证明:dS?I(t)????1?x2?y2(1?x?y)

23221?x2?y2dS???

dxdy---------------------------2 221?x?yx2?y2?tr(1?r2)?2??dr,t?(0,??)----------------------------------------2 01?r2

?(1?t)?0-----------------------------------------------1 令I?(t)?1?t

易知唯一驻点t?1亦为其极大点--------------------------------------1 21r(1?r)------------------------------------------------1 则maxI(t)?I(1)?2??01?r2t?(0,??)

r2?u11?u???du-------------------------------------------------------------1 01?u

?(2ln2?1)?,故原题得证.-----------------------------1

例17(10分)设?是由锥面z?x2?y2与半球面z?R2?x2?y2围成的空间区域,

??是?的整个边界的外侧,求I???xdydz?ydzdx?zdxdy.

解: 由Gauss公式,得I???xdydz?ydzdx?zdxdy?3???dv-----------------------------2

??

因为?在xoy平面的投影区域为

x?y?有3222 ----------------------------------------------2

???dv?3??2?0d?0r-----------------------------------------------------------------2

?6? 0r)rdr??R3(2?2).--------------------------------------------2

例17(8分)求I????f?x,y,z??x?dydz??2f?x,y,z??y?dzdx??f?x,y,z??z?dxdy,其中

Σ

f?x,y,z?为连续函数,Σ是平面x?y?z?1在第四卦限部分的上侧.

?1解:设Σ的单位法向量n0??cos?,cos?,cos????1,?1,1?,---------------------------------2 则I??f?x,y,z??x??cos????2f?x,y,z??y??cos????f?x,y,z??z??cos??dS------1 ??

??

Σ

????f?

x,y,z?f?

x,y,z??f?

x,y,z??dS???

dS---------1 ?ΣΣ?

x?y?1?x?y?σ----------------------------------------------------------------1 Dxy

其中Dxy??x,y0?x?1,x?1?y?0--------------------------------------------------------------1 故I???

Dxy??dxdy?1.------------------------------------------------------------------------------------1 2

例18(8分)计算曲面积分I?Σxdydz?ydzdx?zdxdy

Ω??x,y,z?x?2,y?2,z?2边界曲面的外侧.

解:命P????xy2?y2?z322?,其中Σ为空间区域x

?x2?y2?z322?,Q??x2?y2?z322,R?z?x2?y2?z322。

作辅助曲面Σ1为球面x2?y2?z2?ε2的外则,其中 0 <ε< 1。则

I?

Σ?Σ1xdydz?ydzdx?zdxdy?x2?y2?z

22322???Σ1xdydz?ydzdx?zdxdy?x2?y2?z322,----------------------------2 其中Σ?Σ1???xdydz?ydzdx?zdxdy322

2?x?y?z?3?x?y?z??3x?3y?????x?y?z?222?122522??P?Q?R?????????dxdydz ?x?y?z?Ω1?2?3z2dxdydz?0 --------------------------------3

(Ω1为Σ与Σ1之间的空间区域)。所以

I???Σ1xdydz?ydzdx?zdxdy?x2?y2?z3

22??1

ε3??xdydz?ydzdx?zdxdy

Σ1

?11433dxdydz??3?πε?4π.--------------------------------------------3 3??3εΣ1ε3

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