微积分(大学数学基础教程答案)大学数学基础教程(二)多元函数微积分王宝富 钮海第二章习题解答(下).doc
习题2-1 1、解:在任意一个面积微元上的压d,力微元,所以,该平面薄片一侧所受的水压力dF,,gxd,
F,,gxd,,,D
2、解:在任意一个面积微元上的电荷微元,d,dF,,(x,y)d,所以,该平面薄片的电荷总量 Q,,(x,y)d,,,D
223、解:因为,所以,又为单lnu0,x,1,0,y,1x,y,1,x,y,1
22调递增函数,所以,由二重积分的保序,,,,lnx,y,1,lnx,y,1
性得
22,, ln,,x,y,1d,,lnx,y,1d,,,,,0,x,10,x,1
0,y,10,y,1
4、解:积分区域D如图2-1-1所示,所以该物体的质量
12,y1884222223 ()()(44)M,x,yd,,dyx,ydx,,y,y,ydy,,,,,,0y0333D
5、解:(1)积分区域如图2-1-2所示,所以
y111 dyf(x,y)dx,dxf(x,y)dy,,,,x000
(2)积分区域如图2-1-3所示,所以224yx dyf(x,y)dx,dxf(x,y)dy2,,,,00/2yx
(3)积分区域如图2-1-4所示,所以
2222x,x11,y,1dxf(x,y)dy,dyf(x,y)dx ,,,,12,x02,y
(4)积分区域如图2-1-5所示,所以elnx1e dxf(x,y)dy,dyf(x,y)dxy,,,,000e
6、解:(1)积分区域如图2-1-6所示,所以
11x122416,,3/4311/45 ,,xyd,,dxxydy,xx,xdx,x,x,,,2,,,,,0x03311555,,D0(2)积分区域如图2-1-7所示,所以
224,y21642222 xyd,,dyxydx,y,ydy,2(4),,,,,2,00215D
(3)积分区域如图2-1-8所示,所以
01,x11,x0x,yx,yx,yx1,x,1,xx1,x,1,x,ed,dxedy,dxedy,e(e,e)dx,e(e,e)dx,,,,,,,,,1,1,x0,1,x,1D 012x,1,12x,1,(ee,e)dx,(e,ee)dx,e,e,,,10
(4)积分区域如图2-1-9所示,所以
2y219313,,222232 ()()x,y,xd,,dyx,y,xdx,y,ydy,,,,,,,,0y/202486,,D
7、解:
(1)积分区域如图2-1-10所示,令,所以x,rcos,,y,rsin,,,,,,,0,r,a,故 ,22
,a2 ,,fx,yd,,d,r,f(rcos,rsin)dr,,,,,0,D2
(2)积分区域如图2-1-11所示,令,所以x,rcos,,y,rsin,
,故 0,,,,,0,r,2sin,
,2sin, f(x,y)d,,d,r,f(rcos,,rsin,)dr,,,,00D
8、解:
(1)积分区域如图2-1-12所示,令x,rcos,,y,rsin,,所以
,,sin,0,,,0,r,,故 24cos,
1,sin,,,,1x222,12444cos,,,dx(x,y)dy,,dr,rdr,sec,tan,d,,sec,,2,1 02,,,,,0000x
(2)积分区域如图2-1-13所示,令x,rcos,,y,rsin,,所以
,故 0,,,,,0,r,2sin,
,422aa,ya,a223()2,,, dyxydxd,rdr,,,,80000
9、解:(1)积分区域如图2-1-14所示,故
22x219x23() d,,xdxdy,,x,xdx,122,,,,,114yyxD
(2)积分区域如图2-1-15所示,令,所以x,rcos,,y,rsin,
,0,,,0,r,1,故,2
,,2222111,x,y1,r1,r,,2d,d,rdr,rdr222,,,,,000421,x,y1,r1,rD
3,11,,rr,,,dr,dr,,,,004421,r1,r,,
24,11,,1dr1d(1,r),,,,,,,,00442241,r1,r,,
11,,1,,11242,,,,,arcsinr,(1,r),,,202228,,0,,(3)积分区域如图2-1-16所示, 故
3aya33a2222224(x,y)d,,dy(x,y)dx,(2ay,ay,)dy,14a ,,,,,,ayaa3D
(4)积分区域如图2-1-17所示,令,所以x,rcos,,y,rsin,
, 0,,,2,,a,r,b
12,b2,222332故 (xy)ddrdrba,,,,,,,,,,,,0a3D
10、解:积分区域如图2-1-18所示,由图形的对称性得:S,4S,4d,,所以 1,,D1
,,,asin2,222444S,4,drdr,2asin2,d,,,acos2,,a ,,,0000
图2-1-1 图2-1-2 图2-1-3 图2-1-4
图2-1-5 图2-1-6 图2-1-7 图2-1-8
图2-1-9 图2-1-10 图2-1-11 图2-1-12
图2-1-13 图2-1-14 图2-1-15 图2-1-16
图2-1-17 图2-1-18 习题2-2
1、解: Q,,(x,y,z)dv,,,,
2、化三重积分为直角坐标中的累次积分
解:(1)因为积分区域的上曲面为开口向上的旋转抛物面,
22,下曲面为,积分区域在坐标面上的投影区z,0,xoyz,x,y
D:0,x,1;0,y,1,x域,所以 xy
2211,,xyx,,,,fx,y,zdv,dxdyfx,y,zdz ,,,,,,000,
2(2)因为积分区域的上曲面为开口向下的抛物柱面,z,2,x
22与下曲面为开口向上的旋转抛物面围成,二曲面的z,x,2y
22交线在平面上的投影为圆,即xoyx,y,1
1x1,,,,
,22:1xy1x,,,,,,,所以,
222,x2yz2x,,,,,
22,,11x2x,,,, fx,y,zdv,dxdyfx,y,zdz222,,,,,,,,,,11xx2y,
(3)因为积分区域,的上曲面为开口向上的旋转抛物面
22z,0,下曲面为,积分区域,在坐标面上的投影区xoyz,x,y
2D:,1,x,1;x,y,1域,所以 xy
2211x,y ,,,,fx,y,zdv,dxdyfx,y,zdz2,,,,,,,x10,
3、解:积分区域如图2-2-1所示 ,
y111111126 xzdxdydz,xdxdyzdz,xdxydy,x(1,x)dx,022,,,,,,,,,,x,x,101126,
另解:因为积分区域关于坐标面对称,又关,f(x,y,z),xzyoz
于第一坐标是奇函数,所以。 xzdxdydz,0,,,,
4、解:积分区域如图2-2-2所示,当时,过作0,z,h,(0,0,z)
2,R,,22,x,y,z,,D:平行与面的平面,与立体的截面为圆,,xoy,zh,,,z,z,
2,RR2zzD因而的半径为,面积为,故 z2hh
222hh,R,Rh3 ,,,zdxdydzzdzdxdyzdz,,,,,,,2004h,Dz
5、求下列立体的体积 ,
解(1)曲面所围立体是球体与旋转抛物面的一部分(如图2-2-3所示),用柱面坐标计算:
,,225,,,2V,dv,rdrddz,,ddrrdzr,,,,,,,,,004,, 32,1122422,[,(5,r),r],,(55,4)0,03163
图2-2-1 图2-2-2 图
2-2-3
(2)因为积分区域的上曲面为平面,下曲面为,z,1,xz,0,
2D:y,x,1;,1,y,1积分区域在坐标面上的投影区域,所,xoyxy以
,x111111118,,24 (1)2V,dv,dydxdz,dy,xdx,,y,ydy,,,22,,,,,,,,,,y,y10102215,,,
6、利用柱面坐标计算下列三重积分 解:(1)因为积分区域的上曲面为开口向上的上半球面,
2222z,2,x,y,下曲面为开口向上的旋转抛物面,将z,x,y
2222z,2,x,y代入得z,2,z,解此方程得积分区z,1z,x,y
22D:x,y,1域在坐标面上的投影区域,由柱坐标公式,xoyxy
D:0,,,2,,0,r,1得: xy
22,12,r117,2422。 ,,zdv,ddrzrdz,r,r,rdr,,,2,,,,,,,00r0212,
(2)因为积分区域的上曲面为平面,下曲面为开口向,z,2
222222上的旋转抛物面,将代入得,z,2x,y,42z,x,y2z,x,y
22D:x,y,4所以积分区域在坐标面上的投影区域,由柱,xoyxy
D:0,,,2,,0,r,2坐标公式得: xy
2222,116,,,22332()22。 x,ydv,ddrrdz,r,rdr,,,,,2,,,,,,,00/20r23,,,
7、利用球面坐标计算下列三重积分 解:(1)用球面坐标计算
,,2122244,,,,,,()sinsinx,y,zdv,rdrdd,ddrdr,,,,,,,,,000,,
114,,,52(cos),,,,,,r,,,,055,,0
(2)用球面坐标计算
,,,2/42cosa23,,,,,,,,zdvrcosrsindrdddsincosdrdr,,,,,,,,,,,,000,,
,,/4/41445,,,,,,,,,2sincos(2acos)d8asincosd ,,,,,,004
,/44,8a764,,,cos,,a660
8、选用适当的坐标计算下列三重积分
,0,r,cos,
,,,解:(1)积分区域为球,故用球面坐标计算:,,,:0,,,2,0,,,2,,所以
,,,,,2/2cos/2cos22223,,,,,,x,y,zdv,ddr,rsindr,2sindrdr,,,,,,,,00000
,/2,,11,/245,,,d,,,,2sincoscos,,,,,,0042510
22(2)将代入得到平面上的一个圆z,2yxoyz,x,y
2,,111x2y22,,x,y,1,1,用直角坐标公式计算,zdv,dxdyzdz222,,,,,,,,,,111xxy,
由于计算量较大,请同学一试。
用柱坐标计算 x,rcos,,y,rsin,,z,z
,,,,,2sin2sin2sinr1224,,,zdv,ddrzrdz,dr(4rsin,r)dr2,,,,,,,,00r002, ,,81653156,sin,d,,,,,03364226
(3)用柱坐标x,rcos,,y,rsin,,z,z计算
2,121r2r116223 2()zdv,d,drzrdz,,rzdr,,,,,,,,,00000315,
(4)用直角坐标计算
212xxyx1111yx232344 xyzdv,xdxydyzdz,xdxxydy,dx,,,,,,,,,,000000428364,
习题2-3
1、 解:(1)因为连接点(1,0)和(2,1)的直线段的
方程为,所以 y,x,1,1,x,2
22,,112 (x,y)ds,[x,(x,1)]1,(1)dx,2dx,2,,,L11
,2nn22222222x,yds,acost,asint(,asint),(acost)dt,,,,,,0L(2) 2,2n,12n,1,adt,2a,,02,222yds,2a1,cost[a(1,cost)],(asint)dt,,0L(3) 332,22,2a(1,cost)dt,4,a,0
33(4)因为星形线的参数方程为,所以 x,acost,y,asint
,222,,22223332,,x,yds,4a(,3acostsint),(3asintcost)dx,,,,0L,, 555,,233322,12acostsintdx,6asint,6a,00
(5)因为折线ABCD由线段AB,BC和CD构成,在线段AB上,x,y,0,在线的BC上,y,0,而在线段CD上,x,1,z,2,y,t,0,t,3且ds,dy,dt
322222 xyzds,xyzds,xyzds,xyzds,0,0,1,t,2dt,9,,,,,L0ABBCCD
t022,,zds,tcost,tsint,(sint,tcost),1dt,,L0
3 (6)3,,t11022222,,2td(2t)2t2,,,,,,,,0,023,,
1r,2、解:因为曲线L的极坐标方程为,所以 ,
442,1,2233,,,s,ds,r,rd,,d,,又 332,,,L,44
,2,,utan1cos1(sin),udu,d,,du,,,,,22222sinucosusinu(1,sinu)
11/21/2,,21,,,,,d(sinu) ,,,,21,sinu1,sinusinu,,
111sin,u,,,ln,Csinu21,sinu
22,,,111,,,,,,ln,C2,21,,,,
42,,1533s,d,,,ln所以 32,122,4
,,,,2aa222,,,s,ds,r,rd,e,aed,,,,,L00
3、解: 2,1,a,aa,2,1,aed,,(e,1),0a
习题2-4
221、(1)解:将曲面向xoy平面投影,得投影区域D:x+yxy
2?R,从而有 R222zdS,R,x,y,dxdy,,,,222DxyR,x,y,
23,Rdxdy,R,,R,,R,,Dxy
(2) 解:将平面向XOY平面投影,得投影区域,
,
0,x,2,D:,从而有积分 xy3x0,y,3,,442xy(4,2x,y,2x,y)1,z,zdxdy4(z,2x,y)dS,3322,,,,3Dxy
,
461,dxdy,461,,3Dxy
22,:z,1(3)解:由 ()得 x,y,11
22dS,dxdy,D:x,y,1 xy
1222 由 得 ,:z,(x,y)(0,z,1)2
122xy2dS,(1,,)dxdy,2dxdy2222x,yx,y
22D:x,y,1xy21,12,222222;zx,ydS,x,ydxdy,,drdr,,,,,,,,,,33,D001xy 212,22222()22zx,ydS,x,ydxdy,d,rrdr,,,,,,,,002,D2xy
,22222222zx,ydS,(x,y)dS,(x,y)dS,,,所以, ,,,,,,32,,,12
2、解:将被截得的平面向XOY平面投影,又有已知条件的,
ccz,c,x,y, ab
ccz,,,z,,,设所求的面积为A,则有 xyab
cc122222222AdS1()()dxdyabaccb ,,,,,,,,,,,,,ab2,Dxy
3、解:将曲面向XOY平面投影,得投影区域,
22D:x,y,4,且z,,2x,z,,2y, xyxy
设所求的面积为A,则有
,,,,,,AdS1(2x)(2y)dxdy,,,,22
,Dxy ,,1,3,,,,d,14rrdr,(17)1,,22,,226,,00
4、解:以圆环的中心为坐标原点建立坐标系,则容易知道
圆环薄片的面密度为:
122,设薄片的质量为M,则有 ,(x,y),,当x,y,4时22x,y
24,11 M,,2,2,dxdy,4,,d,,rdr,8,,,,,0222rx,y,
2aaa222,?,k,k,,,,,(,),,5、,而 ?(x,y),k(x,y),0002222a
aa24a,2222220()() ,,,,,Mxydsdxxydy,,0022,,,,3aa00s
习题2-5
x133,,,xxyd,xyd,xydydx(,)()2,,,,,,x0DD
1、 解: 68xx111,,,()|026848
x12222(,),,()y,xyd,xyd,xydydx2,,,,,,x0DD
69xx111,,,()|036954
x122(,),,(),xyd,xyd,xydydx2,,,,,,x0DD
57xx111,,,()|025735
11
,,353535354854 x,,,y,,,重心(,).1148544854
3535
2、 解:设P(x,y)为三角形上一点,则容易知道此点的密度
22为。 ,(x,y),x,y
axa2222(,)()(()),,,,,xxydxxydxxydydx,,,00,,,,,,DD 54233254xaxaxaxaa()|,,,,,,01523615
aya2222,(,)()(()),,,,yxydyxydyxydxdy,,,,,,,,,00DD
54233254yayayayaa()|,,,,,,01523615
axa,2222(,)()(()),,,,xydxydxydydx,,,,,,,,,00DD
43223442xaxaxaxaa()|,,,,,,0123236
2a2a(,)重心: 55
3、 解:(1)由对称性知道重心一定在z轴上。
1122,zdv,zdv,(zdz)dxdy,[1,(x,y)]dxdy22,,,,,,,,,,,,xy2,,DDxyxy
24,,212,rr1121,,rrdrd,,d,((1))()|,,0,,,00022244
,3V,(0,0,)而圆锥的体积为:。所以重心为:。 34
(2)容易知道此几何体是两个同心半球之间的部分,且重心一定在z轴上。而
,,2A2,,,,,,,dv,dv,dd,sind,,,,,,,,,00a,,
,3,,2A332,2,[(,cos)|],(|),(A,a),,0a33
,,2A2,,,,,,,,,zdv,zdv,ddcos,,sind,,,,,,,,,00a,,
,4,,12A442,2,(sin|),(|),(A,a),,0a244
443(A,a)重心:(0,0,)。 338(A,a)
4、 解:以圆柱下底面的圆心为坐标原点,以转动轴为z轴建立坐标系,设P(x,y,z)为圆柱体上一点,则此点到
22r,x,y转动轴的距离为,因此
2h,a2222,,Ir(x,y,z)dv(xy)dvdzdrrdr,,,,,z,,,,,,,,,000,, 14ha,,2
5、 解:设P(x,y,z)为弹簧上一点,则P到Z轴的距离为
22r,x,y,因此有
,22222222,I,r,(x,y,z)ds,a,(a,kt),a,kdtz,,L0 122222232,222223,aa,k,(at,kt)|,aa,k(3a,4k),,033
6、 解:有对称性知道F=0。 y
,,,R(x,y)xrcos,,,22()F,Gd,Grdrdx,,,,,33R,122222D222(x,y,a)(r,a)
2R1a2,2[],G,dr,13R1222222(r,a)(r,a)
2rrrRR22,2[(ln(1)|)(|)],G,,,RR21122aaa,r
22R,R,aRR22212,[ln],G,,222222R,R,aa,Ra,R1121
,,R(x,y)12,,,2()F,,aGd,,aGrdrdz,33,,,,R,122222D222()()x,y,ar,a
R11222,,,,aGd(r,a)3,R12222(r,a)
1R2,,,aG|R122a,r
11,,(),aG,2222a,Ra,R21
7、 解:由对称性知道F=F=0。 xy
,
(x,y,z)(z,a)F,Gdvz,,,3,2222[x,y,(z,a)]
,,,Rh2(z,a),Gdrdrdz,,,3000222[r,(z,a)]
1,R,,h222,,2G{[(z,a),r]|}rdr0,0
11,,R2222,,22 ,,2G{[(h,a),r],[a,r]}rdr,0
11R222222,,,,2G{[(h,a),r],(a,r)}|0
11222222,,,,2G{[(h,a),R],(a,R),[(h,a),a]}
11222222,,2,G,{[(h,a),R],(a,R),h,2a}