[理学]高数 第十一章 曲线积分与曲面积分
第十一章 曲线积分与曲面积分
?1 对弧长的曲线积分
计算公式:无论是对弧长还是对坐标的曲线积分重要的是写出曲线的参数方程
xxt,,,,b22,'',,,,Latb:,,,则 若fxydsfxtytxtytdt,,,,,,,,,,,,,,,,,,,,,,,Layyt,,,,,
,xxt,,,
,若,则 Lyytatb:,,,,,,
,zzt,,,,
b222''',,,,,,fxyzdsfxtytztxtytztdt,,,,,,, ,,,,,,,,,,,,,,,,,,,,,,,,La
注意:上限一定要大于下限
1( 计算下列对弧长的曲线积分
222222L),其中为圆周; (1(x,y)dsx,y,a,L
22222解:法一:()xyds,,()ads ,,LL
445,ads ,,aaa(2)2,,,L
xa,cos,,法二:, L:02,,,,,ya,sin,,
222()xyds, ,L
2,22222,,,,[cossin]sincosaaaad,,,,, ,,,,,,,,,0
2,55,,ada,,2 ,0
22x,y222Ledsy,x(2),其中为圆周,直线及轴在第一象限内所围成的x,y,ax,L
B
A 扇形的整个边界;
2222xyxy,,edseds,,,()解:,其中 ,,,,LOAABBO
xx,xa,cos,xx,,,,,2,, OAxa:,0,,AB:,0,,,BOxa:0,,,,,y,0ya,sin4yx,,2,,,
a222,xy,022xedsedx,,10 ,,0oA
axa,,,edxe1 ,0
a22,aexy,aaedseds,,, eds,,,ABABAB4
22xy,eds(或 ,AB
,2222aacossin,,,,,,,4 ,,,eaadsincos,,,,,,,,0
,a,aea4,,ead) ,,04
2a2222,xy,xx222 edsedx,,11,,BO0
2axa22 ,,,edxe21,0
22,xy,aedsea,,,(2)2故 ,L4
2L(3),其中为抛物线上介于与之间的一段弧; y,2x,1xdsx,0x,1,L
xx,,解:由,得 Lx:01,,,2yx,,21,
12 xdsxxdx,,14,,,,L0
32212(116),x017171,3,, 3248
2Lx,a(t,sint),y,a(1,cost)(0,t,2,)(4),其中为摆线的一拱; yds,L
2,2222ydsatatatdt,,,,(1cos)[1cos]sin,,,,,,,,解:L 0
5,232 ,,2(1cos)atdt,0
5,22,ttt35322,2(2sin)adt,8sinadt(令,,) ,,00222
,35,16sinad,, ,0
,4225635332,,,,,,, 32sin32adaa,05315
222L(5),其中为圆周; xydsx,y,a,L
xa,cos,,,L:0,,,xydsxyds,4解:利用对称性,其中 ,1,,LL1ya,sin2,,
xydsxydsxyds,,44,,,LLL11
,222 ,,,4(cos)(sin)(sin)(cos)aaaad,,,,,,0
,,332322 ,,,4cossin2sin2adaa,,,,0,0
1ttt,t(6),其中为曲线x,ecost,,z,e上相应于从0变到dsy,esint,,222,,xyz
2的弧段;
解:
1ds 222,,,,xyz
2''1ttt222 ,,,etetedt[cos][sin],,,,222,0tttetete,,cossin,,,,,,
233,,t2,,,edte(1) ,022
222,x,y,z,2,,,:yds(7),其中为空间圆周: . ,,,,y,x,
222x,cos,,,xyz,,,2,22解:由,得,令 22xz,,02,,,,,,yx,z,2sin,,,,
,x,cos,
,故。故 ,,,,:cos02y,,,,
,z,2sin,,
yds ,,
2,222,,,cossinsin2cos,,,,,d ,0
2,,2cos,,d ,0
,,32,22 ,,,,2[coscoscos]42,,,,,,ddd3,,,,,022
x,acost,,2( 螺旋形弹簧一圈的方程为: ,设它的线密度为y,asint (0,t,2,),
,z,kt,
222,求: ,(x,y,z),x,y,z
z(1) 它关于轴的转动惯量;(2)它的重心坐标. Iz
22Ixyds,,, (1),,z,L
22222,,,,xyxyzds ,,,,,L
2,2,222222222222,,,aaktakdt,,,aakaktdt ,,,,,,00
2222222,,,,, (34)aakak 3
222xxyzds,,,,,L(2) x,222xyzds,,,,,L
2,22222ataktakdtcos,,,,,0 ,2,22222aktakdt,,,,,0
2,2222aktatdt,cos,,6ak,0(分子采用分部积分法) ,,2222,222,34ak,aktdt,,,,0
222yxyzds,,,,,L y,222xyzds,,,,,L
2,22222ataktakdtsin,,,,,0 ,2,22222aktakdt,,,,,0
2,6 ,ak ,22234ak,,
222zxyzds,,,,,L z,222xyzds,,,,,L
2,22222ktaktakdt,,,,,0 ,2,22222aktakdt,,,,,0
2223(2),,kak,= 22234ak,,
?2 对坐标的曲线积分 无论是对弧长还是对坐标的曲线积分重要的是写出曲线的参数方程
,xxt,,,,Lt::,,,1计算公式:若,(其中分别始点和终点对应的参数),则 ,,,,yyt,,,,,
,''PxydxQxydyPxtytxtQxtytytdt,,[,,],,, ,,,,,,,,,,,,,,,,,,,,,,L,
xxt,,,,
,若,(其中分别始点和终点对应的参数),则 ,,,Lyytt::,,,,,,,
,zzt,,,,
PxyzdxQxyzdyRxyzdz,,,,,,,, ,,,,,,,L
,''',,,[,,,,,,]PxtytztxtQxtytztytRxtytztztdt ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
注意:(1)对定向曲线才能说对坐标的曲线积;定向曲线的参数方程与未定向曲线的参数
方程的不同:
? 定向曲线的参数
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
示为始点的参数到终点的参数而不管谁大谁小: t:,,,
? 未定向曲线的参数方程的参数表示为不等式: atb,,(2)?弧长的积分转化为定积分时定积分的上限一定要大于下限 ?对坐标的曲线积分转化为定积分时定积分的上限一定是终点的参数,下限是始点的
参数,而不管上限是否一定要大于下限
2:两类曲线积分的关系
(1) 定向曲线的切向量及其方向余弦
,xxt,,,,Lt::,,,若 ,yyt,,,,,
?当时 ,,,
''xtyt,切向量为:; ,,,,,,
''xtyt,,,,方向余弦为 cos,cos,,,,2222''''xtytxtyt,,,,,,,,,,,,,,,,,,?当时 ,,,
'',,xtyt,切向量为:; ,,,,,,
''xtyt,,,,,,方向余弦为 cos,cos,,,,2222''''xtytxtyt,,,,,,,,,,,,,,,,,,类似可以推广到空间曲线。
(2) 两类曲线积分的关系
PxydxQxydyPxyQxyds,,[,cos,cos],,,,, ,,,,,,,,,,LL
其中为定向曲线切向量的方向余弦 cos,cos,,
注意:把第二类曲线积分转化为第一类曲线积分其关键是求出切向量。特别要注意始点
参数与终点参数大小关系对切向量符号的影响。
L1( 把对坐标的曲线积分化为对弧长的曲线积分,其中为: P(x,y)dx,Q(x,y)dy,L
2(1)从点(0,0)沿抛物线到点(1,1); y,x
xx,,xy,1,2x解:,由,故在处切向量为,所以 01,Lx::01,,,,,,2yx,,
11, ,,,cos,22,14x,12x,,
22xx所以 ,,,cos22,14x,12x,,
PxydxQxydy,,, ,,,,,L
,,[,cos,cos]PxyQxyds,, ,,,,,L
PxyxQxy(,)2(,),,ds ,2L14,x
22y,0(2)从点(0,0)沿上半圆周x,y,2x到点(1,1).
xx,,,,1,x,xy,解:,由,故在处切向量为,所以 01,Lx::01,1,,,,,,22yxx,,22xx,,,,,
1,x
212xx,2,所以 ,,,,,cos2,xxcos1,,,x22,,,,1,x,1x1,,1,,,,222xx,,2xx,,,,
PxydxQxydy,,,,,,,,L
,,[,cos,cos]PxyQxyds,,,,,,,L
2 ,,,,[2(,)(1)(,)]xxPxyxQxyds,L
(或,,,[(,)(1)(,)]yPxyxQxyds) ,L
x,,1cos,,,,,法二,由, L:,:,,,,,2y,sin2,,
,,,(sin),cos,,sin,cos,,,故切向量为,即 ,,,,所以
sin,, ,,,cossiny,,22,,aasincos,,,,,,
,cos,,所以 coscos1,,,,,x,,22sincos,,,,,,,,
PxydxQxydy,,,,,[,cos,cos]PxyQxyds,, ,,,,,,,,,,LL
,,,[(,)(1)(,)]yPxyxQxyds ,L
2( 计算下列对坐标的曲线积分:
222L(1),其中为抛物线上从点(0,0)到(2,4)的一段弧; (x,y)dxy,x,L
xx,,解:由,得 Lx::02,,2yx,,
22562222xydxxxdx,,,,,()() ,,,,L015
222L(2),其中为圆周及轴所围成的在第一象限内的区域(x,a),y,a(a,0)xy dxx,L
的整个边界曲线弧(按逆时针方向);
xydxxydx (),,解:, ,,,LOAAO
a
xx,,OAxa::02,其中, ,y,0,O A
xaa,,cos,, AO::0,,,,ya,sin,,
(注意此方程不是的极坐标方程,故不能说在极坐标系下的范围,事实上极坐,,:0,,
,,ra,,2cos,:0:0,标方程为,故在极坐标系下的范围为) ,,,22
2axydxxdx,,,00 ,,OA0
,xydxaaada,,,cossincos,,, ,,,,,,AO0
,322,,,adsinsincos,,,, ,,,0
,,32222 ,,,add[2sinsincos],,,,,,,00
3a,,3,,,,,a(0) 22
33,,aa 0(),,,,,故xydx ,L22
222L(3),为从点(1,0)到点(-1,0)的上半椭圆周; (1,2xy)dx,xdyx,2y,1(y,0),L
x,cos,,,解:由,得 L::0,,,,2y,sin,,,2
2(12),,xydxxdy ,L
,222,,,,,,,,,,[12cos(sin)](sin)coscos]d ,022
,,,223,,,,,,,,,,,sin2sincoscosddd ,,,0002
,,222,,,,,(1sin)sind,,cos2sinsind,,, 0,,002
33,,sin,2sin,,,,,,22 ,,sin,0,,0323,,
,,,,,,2002
x,ydx,x,ydy()()222L(4),其中为圆周x,y,a(按逆时针方向); ,L22x,y
xa,cos,,解:由,得 L::02,,,,ya,sin,,
()()xydxxydy,,, 22,Lxy,
2,(cossin)(sin)(cossin)cosaaaaaa,,,,,,,,,, ,d,2,0a
2,,,,,d,,2 ,0
22,x,y,1,,z),其中为椭圆周:,且从轴正方(5(z,y)dx,(x,z)dy,(x,z)dz,,,,x,y,z,2,
,向看去,取顺时针方向;
x,cos,,22,,x,y,1,,,,:sin:20y解:由 得,故 ,,,,,,x,y,z,2,,z,,,2cossin,,,
()()()zydxxzdyxzdz,,,,,,,
022,,,,,[(4cossin)4(cossin)cossin,,,,,,,d ,,2
,,,,,,3003,,
22,,22cossin,,,,dd,(注意:易知,所以 ,,00
22,,22cossin,,, ,,00
22,,1122,,,,,,,d,,(cossin)d ,,0022
x,t,
,2222,t(6),其中是曲线:,上由0到的一段弧. yt(y,z)dx,2yzdy,xdz2,,,,,3z,t,
222()2yzdxyzdyxdz,,,解: ,,
2,643846457,,,,,,,32ttdt ,,,057
2L3(计算,其中:(1)抛物线上从点(1,1)到点(4,2)y,x(x,y)dx,(y,x)dy,L
的一段弧;(2)从点(1,1)到点(4,2)的直线段;
22(3)曲线x,2t,t,1,y,t,1上从点(1,1)到点(4,2)的一段弧.
2,xy,解:(1)由,得 Ly::12,,yy,,
()()xydxyxdy,,,,L
23422,, ,,,,,,()2()yyyyydy,,,13
xx,,,(2)由,得 Lx::14,12,yx,,,33,
()()xydxyxdy,,, ,L
412121,,,,,,,,,()()11xxxxdx ,,,133333,,
2,xtt,,,21,(3)由,得 t:01,,2yt,,1,,
()()xydxyxdy,,, ,L
13222,,,,,,,,,,(32)41(2)2ttttttdt ,,,,,03
22L4(证明: 其中为平面上光滑曲线的长度. sin(x,y)dx,cos(xy)dy,2ll,L
(提示:转化为对弧长的曲线积分) 证明:
22sin()cos()xydxxydy,, ,L
22,,,[sin()coscos()cos]xyxyds,, ,L
22其中是切向量的方向余弦,故满足。 cos,cos,,coscos1,,,,
2222sin()cos()xydxxydy,,,,,sin()coscos()cosxyxyds,, ,,LL
222 ,,,[(sin()coscos()cos)]xyxyds,,,L
22222222 ,,,,,(sin()cos2sin()coscos()cos)cos()cos)xyxyxyxyds,,,,,L
222222222222,,,,,,(sin()cos[sin()coscoscos()]cos()cosxyxyxyxyds,,,,,L
22222222 ,,,,,(sin()[coscos]cos()[coscos]xyxyds,,,,,L
2222 ,,,,,(sin()cos()22xyxydsdsl,,LL
法二:证明:
22sin()cos()xydxxydy,, ,L
22,,,[sin()coscos()cos]xyxyds,, ,L
22其中是切向量的方向余弦,故满足。 cos,cos,,coscos1,,,,
2222sin()cos()xydxxydy,,,,,sin()coscos()cosxyxyds,, ,,LL
22n,cos,cos,,nxyxy,,sin(),cos()设向量,则 ,,,,e22sin()coscos()cosxyxynn,,,,,, e
2222,nn,,,sin()cos()xyxy, ,2e
2222sin()cos()xydxxydy,,,sin()coscos()cosxyxyds,,,,故 ,,LL2ds,2l ,L
?3 Green公式
1( 用曲线积分计算下列曲线所围平面图形的面积:
22yx,,1(1)椭圆:; 22ab
xa,cos,,解:若:,则 L::02,,,,yb,sin,,
1Adxdyydx,,,, ,,,DL2
2,122,,,,,,,,,ababdabcossin ,,,02
33(a,0,0,t,2,)(2)星形线:,. x,acost,y,asint
3,xatcos,,解:若:,则 ,Lt::02,,3yatsin,,,
1Adxdyydx,,,, ,,,DL2
2,1242242,,,,3cossin3sincosattattdt ,,,02
22,3a42242,,,, cossin3sincosttattdt,,,02
22,3a22, sincosttdt,02
22,3a2, sin2tdt,08
22,31cos43at,2, ,,dta,0828
2(用格林公式计算下列曲线积分
22L(a,0)(1),其中为圆周,取逆时针方向; xydy,xydx,L
xLD:0,x,,,0,y,sinx(2),其中为闭区域的正向边e[(1,cosy)dx,(y,siny)dy],L
界.
,,QP2222PxyQxyxy,,,?,,,,,解:(1), 222x,y,a,,xy
222L又逆时针方向,设,所以 Dxya:,,2222xydyxydxxyd,,,, ,,,,,LD
2a,124,,drrdra,, ,,002
22222xydyxydxxydad,,,,,,(注意,为什么,) ,,,,,,,LDD
,,QPxx(1cos),(sin),PeyQyyye,,,,,?,,,(2) ,,xy
yx,sin
xe[(1,cosy)dx,(y,siny)dy]所以 ,L
,sinx, xx,,,,yeddxyedy, ,,,,D00
,sinxx,,edxydy ,,00
,1x2,,exdxsin ,02
,11cos2,xx,,edx ,022
,,1xx,,,edxexdx[cos2] ,,004
111,,,,,,,,, (1)1(1)eee,,4205
,,xxx2,(其中exdxexexdxcos2cos22sin2,, 0,,00
,xx2,,,,,,eexexdx12[sin22cos2] 0,0
,x,,,,eexdx14cos2 ,0
,1x,exdxe,,所以cos21) ,,,05
xdy,ydx222L3(计算积分,其中为圆周(按逆时针方向); (x,1),y,R(R,1),L224x,y
,,,yxQPPQ,,?,,,,0解 222244xyxyxy,,,,
,yx222PQ,,,(1)故当时,在所围的区域R,1(1)(1)xyRR,,,,222244xyxy,,
xdyydx,00,,,d内有连续偏导,满足格林公式条件。 D22,,,LD4xy,
222(2)故当时,所围的区域含有点,故R,1D(0,0)(1)(1)xyRR,,,,
,yx?,,PQ,在区域有点没有连续偏导,不满足格林公式条件。不能直D222244xyxy,,
接用格林公式条件。
222做曲线(取得足够小保证含在所围区域)方向为逆时针,即L,llxy:4,,,
1,,,x,cos,。 l:02,,,2,
,y,sin,,,
,则曲线围成复连通区域且为的正向边界。 DDLl,11
xdyydx,故在复连通区域满足格林公式条件,故 D,122,Ll,4xy,xdyydx,00,,,d即 ,22,,,LlD,14xy,
xdyydxxdyydxxdyydx,,,,,, ,222222,,,Lll444xyxyxy,,,
112222,,,,cossin,2,22,d ,2,0,
2,1,,d,, ,02
222222(注之所以取曲线是方便计算,若取则计算麻烦) lxy:4,,,lxy:,,,
xoy4(证明下列曲线积分在面上与路径无关,并计算积分.
(3,4)2322(6xy,y)dx,(6xy,3xy)dy(1) ,(1,2)
2322xoy解:,所以单连通区域面有连续偏导,且 PxyyQxyxy,,,,6,63
,,QP2,,,123xyxoy,所以曲线积分在面上与路径无关。,,xy
C(3,4) A(1,2)
B(3,2)
(3,4)2322(6xy,y)dx,(6xy,3xy)dy法一: ,(1,2)
2322,,,,,()(6)(63)xyydxxyxydy ,,ABBC
xx,x,3,,其中 ABx::13,BCy::24,,,y,2yy,,,342322,,,(622)xdx,,,,,,,(6333)236yydy ,,12
23223,,,3xyxyy,uxyxyydx(,)(6),,法二设: ,,,
dy,dy,,,,,u,2222则63xyxy得,0 ,,,,,63xyxyydydy,
223,故 uxyxyxyC(,)3,,,
(3,4)2322(6)(63)xyydxxyxydy,,,,,,uu(3,4)(1,2)236 ,(1,2)
(2,1)423(2)(2xy,y,3)dx,(x,4xy)dy ,(1,0)
423解:,所以单连通区域xoy面有连续偏导,且 PxyyQxxy,,,,,23,4
,,QP3,,,24xyxoy,所以曲线积分在面上与路径无关。 ,,xy
法一:
C(2,1) (2,1)423(23)(4)xyydxxxydy,,,, ,(1,0)A(1,0) B(2,0)
423,,,,,,()(23)(4)xyydxxxydy ,,ABBC
xx,x,2,,其中 ABx::12,BCy::01,,,y,0yy,,,21423,,,,(2003)xdx,,,,,(242)5ydy ,,10
424uxyxyydxyxyxyxy(,)(23)3,,,,,,,,,,法二设: ,,,,,
dy,dy,,,,,u,2323,得0,所以 xxyxxy44,,,,,,ydydy,
24, uxyxyxyxC(,)3,,,,
(2,1)423(2xy,y,3)dx,(x,4xy)dy故= uu(2,1)(1,0)5,,,(1,0)
5(用适当的方法计算下列曲线积分
2222L(R,0)(R, 0)(1),其中为圆周上从点(xsin2y,y)dx,(xcos2y,1)dyx,y,R,L
B (0, R)依逆时针方向到点的弧段; A
,,QP2,,1解:由 ,有 PxyyQxy,,,,sin2,cos21D ,,xy
2(sin2)(cos21)xyydxxydyd,,,,, O A ,,,OALBOD,,
xx,x,0,,其中OAxR::0,, BOyR::0,,,y,0yy,,,
2(sin2)(cos21)xyydxxydy,,, ,L
2,,,,(sin2)(cos21)xyydxxydy,OALBO,,
2,,,,,[](sin2)(cos21)xyydxxydy ,,OABO
2 ,,,,,,dxyydxxydy,[](sin2)(cos21),,,,DOABO
2R0,R,,,,,,(sin(20)0)(0cos21) xdxydy,,0R4
220,,RR,,,,,,0(0cos21) ydyR,R44
ydxxdy,L(2, 1)(1, 2)(2),其中为从点到点的直线段. 2,Lx
,,QPy1B(1,2) PQ,,,,,,0解:由 ,有 2xx,,xy
积分与路径无关,则 A(2,1)
ydxxdyydxxdy,,[],, 22,,,C(1,1) LACCBxx
xx,x,1,,其中, ACx::21,CBy::12,,,y,1yy,,,
12ydxxdyydxxdy,,dxdy,3[],,,,,, 2222,,,,,LACCB21xxx12(注意:若应用积分与路径无关,则必须保证在添加的曲线与原曲线所围的区域是单连通的,
和在区域有连续偏导数,如该题中区域就不能含原点) PQ,
6(解下列全微分方程
3232(1); (x,3xy)dx,(y,3xy)dy,0
,,QP3232,,,6xy解: ,在面有,得方程为全微分方程。 xoyPxxyQyxy,,,,3,3,,xy
1332422uxyxxydxy,3,,,,,,,xxyy,法一,故 ,,,,,,,,,42
dy,dy,,,,,u,142323,yy,33xyyxy,,得,即 y,,,,,,,4ydydy,
B(x,y)
1314224xxyyC,,,所以方程通解为 424A(x,0) (,)xy3232uxyxxydxyxydy,(3)(3),,,,法二,令 ,,,(0,0)
O(0,0) 3232,,,,,()(3)(3)xxydxyxydy ,,OAAB
xx,xx,,,其中AByy::0, OAxx::0,,,yy,y,0,,
xy332,,,,,(30)0xxdx,,,0(3)yxydy ,,00
1134422 ,,,xyxy442
1314224所以方程通解为 xxyyC,,,424
xdx,ydy(2). ,xdy,ydx,0221,x,y
,,QPxy,解:,在面有,得方程为全微分PyQx,,,,,xoy2222,,xy11,,,,xyxy
方程。
,,x22,,uxyydxyxyxyy,1,,法一,故 ,,,,,,,,,,,,,,,22,,1xy,,,,
dy,dy,,,uyy,,,xx,y,0,,,,,,得,即 ,0,,2222ydy,dy11xyxy,,,,
22B(x,y) 1,,,,xyxyC所以方程通解为
(,)xyA(x,0) xdxydy,法二,令 ,uxyxdyydx,,,,,,22(0,0)1,,xy
O(0,0) xdxydy, (),,,,xdyydx,,22OAAB1,,xy
xx,xx,,,其中 AByy::0,OAxx::0,,,yy,y,0,,
xyxy,,dx0 0(),,,xdy,,22022010,,x1,,xy
222y,,,,,,,11(1)xxyxy 0
2222,,,,,,,,,,111(10)xxyxyx
22,,,,,11xyxy
2211,,,,,xyxyC所以方程通解为
x,ydx,x,ydy()()L7(计算曲线积分,其中: ,L22x,y
2222(1)闭区域的正向边界; a,x,y,b(b,a,0)
()()xyxy,,,,,PQ,PQ,,,,则 2222,,yxxyxy,,
()()xyxy,,,2222PQ,,,显然在内有连续偏导数,满足格林a,x,y,b(b,a,0)2222xyxy,,
()()xydxxydyQP,,,,,,公式条件,故,,,()0d 22,,,LDxyxy,,,
222(a,0)(2)圆周按逆时针方向; x,y,a
()()xyxy,,,222PQ,,,解:圆周所围区域含原点,故在其内没有连续偏导,x,y,a2222xyxy,,
xR,cos,,数,不能用格林公式。直接计算,故 L::02,,,,yR,sin,,
2,2()()xydxxydya,,,, ,,,,,d2222,,L0xya,
A(,,, ,,)B(,, ,,)(3)从点沿曲线到点的弧段. y,,cosx
(0, ,)E ,,PQ,解:由,则积分路径无关,故: ,,yx(- ,, ,),C (,, ,),D
()()xydxxydy,,,
22,AExy,
(,,-,),B ()()xydxxydy,,,
,(-,,-,)A 22 ,ACCE,xy,
()()xydxxydy,,, 22,EBxy,
()()xydxxydy,,,,, 22,EDDB,xy,
xxxx,,,,,,,,,其中, ACy::CDx::,,DBy::,,,,,,,,,,,,,yyyyy,,,,,,,
x,ydx,x,ydy()()故: ,L22x,y
()()xydxxydy,,,,,[] 22,,AEEBxy,
()()xydxxydy,,,,,,,[] 22,,,,ACCEEDDBxy,
()()xydxxydy,,,,,,[] 22,,,ACCDDBxy,
,,,,()()xydxxydy,,,,,(),y,,,(),y()x,,,dy,dx,dy 2222222,,,,2,,L,,,x,,yxy,,,,,y,,,
,,,,,,y,,yx,,x,,,dy,dx,dy,3dx 22222222,,,,,,,,,,,,xx,,,y,y,,,,
,,,,xx,,,,,3dx,,33dxdx,,06dx 22222222,,,,,,,0,,,x,x,xx,,,,,,
x3,,,,6arctan 02,
8(利用曲线积分与路径无关的条件,求待定参数或函数.
4aa,124(1)确定的值,使曲线积分与路径无关; I,(x,4xy)dx,(6xy,5y)dya,L
,,PQ4124aa,,解:,欲使曲线积分与路径无关当且仅当,即 PxxyQxyy,,,,4,65,,yx
,461aa,,,,
,aa,,122461xayaxy,,,即得 a,3a,,21,,,
,a,,12,
yeI,y,(y)dx,(,,(y))xdy,(y),(1),e(2)求可微函数,,使曲线积分 ,Lyy,0在的开区域内与积分路径无关.
y,,PQe,解:,,,,,(),(()),积分与路径无关当且仅当,即 PyyQyx,,yxy
ydye,(),,,,得 ()()yyy,,dyy
ydyye,,()(),y,,,,(这是以自变量为未知函数的一阶线性微分方程) 0y,,2dyyy
ye,(1),e又得 ,,y,,2y
22,v,v,v,v9(证明的充分必要条件为: dy,dx,0,,0,L22,x,y,x,y
Lv(x,y)其中是单连通开域内的一条简单闭曲线,在内具有连续的二阶偏导数 GG
,,vv,,vv,v,vdydx,PQ,,,,证明:对曲线积分,故的充分必要dy,dx,0,,LL,x,y,,xy,,yx
22,,PQ,,Pv,,Qv,条件为,又, ,,,22,,yx,,xx,,yy
2222,,vv,v,v,v,v故的充分必要条件为, 即 ,,dy,dx,0,,0,22L22,x,y,,yx,x,y
?4 对面积的曲面积分
1(计算下列曲面积分
22(1),其中?为抛物面在面上方的部分; dSxoyzxy,,,2(),,,
2222xyDDxy,,:2,,,解: ,,,,:2()zxy,,xyxy
222222dSzzdxdyxydxdyxydxdy,,,,,,,,,,,1122144则 ,,,,xy
22,222故 ,,drrdr,14dSxy,,,144,,,,,,D00xy,
32,213122222,,,,,,,,,214(41)rdr[14]r ,,,0,08433
2222zxy,,(2),其中?为锥面及平面所围成闭区域的边界曲面. ()xydS,z,1,,,
解:如图,其中 ,,,,,12 ,,:1z2
2222xyDDxy,,:1,,,,,,:,zxy ,,1
2222,,,:zxy xyDDxy,,:1,,,,故 ,,:1,z,,12
222222()xydS,()xydS,()xydS,=+ ,,,,,,,,,12
22,,,,xy22,,,,()1=xydxdy,,, ,,2222,,,,xyxy,,D,,,,
2222+ ()100xydxdy,,,,,D
22 ,,,21()xydxdy,,,,D
21,12,,,,,,21(12)drrdr ,,,,002
2222(3),其中?为锥面被柱面所截zxy,,()xyyzzxdS,,xyaxa,,,2(0),,,
得的部分;
2222xyDDxyax,,:2,,,解: ,,,:zxy,,
22,,,,xy22,,,,,,,1dxdy则 dSzzdxdy,,,1,2dxdyxy2222,,,,xyxy,,,,,,
2222故 ()()2xyyzzxdSxyyxyxyxdxdy,,,,,,,,,,,D,
2222,2[xydxdy ,,yxydxdy,,xyxdxdy],,,,,,DDD
22 ,,,,002xyxdxdy,,D
22yxy,(区域关于轴对称,函数,是关于奇函数) xxyy
,,2cosa2cosa,,322 ,,2cosdrrrdr,,,2cos,,drdr,,,,,,00,,22,452,42cosad ,,,,,2
,426444452,,,,822aa ,82cosad,,,05315
2222zxy,,,4(4)()xydS,,其中?为上半球面. ,,,
2222xyDDxy,,:4,,,,,,,:4zxy解:,则 ,,
22 dSzzdxdy,,,1xy
22,,,,,,xy2,,,,,,,1dxdy ,dxdy222222,,,,44,,,,xyxy4,,xy,,,,
2,222r22222故: ()xydS,,,drdr,,xydxdy(),,,,,,2D0022,4r,,xy4,
22222222,,,4(4),rdr,,24]rrdr ,,,4[4,rr0,,00
3164222,,,,,, 8[4r,,033
3( 计算曲面壳
122,,,,,:()(01)zxyz的质量,面密度. ,,z2
解:质量 MdSzdS,,,,,,,,,
12222xyDDxy,,:2,,,,,,:()zxy其中, ,,2
2222dSzzdxdyxydxdy,,,,,,11 xy
12222则 MzdSxyxydxdy,,,,,()1,,,,D2,
3222,2112222322,,,drrrdr1,,,rdr1 ,,,rrdr1,,,,,,000023
332112222222,,,,,[11]rrrdr ,,,,0,033
32,222,,,,[631(1)]rdr ,,,03
5,2222,,,[631]r ,,035
2,,,(163) 15
222zaxy,,,4( 求密度为常数的均匀半球壳对于Oz轴的转动惯量. ,Iz
22解: IxydS,,(),z,,,
222?在面上的投影区域: Dxoyxya,,xy
2,2ara2222 ,,,IxydS,,(),adrdr,,,()xydxdyz,,,,,,22D00222xy,ar,,axy,
aa222222222a,,,2(),,ardar,,,,,2[],,aarrardr 0,,00
3a24a22222242,,,,2[()],,aardar,,,,,,,,aara 2[],,0,033
?5对坐标的曲面积分 计算联合形式 PdxdyQdydzRdzdx,,,,,
,, 法一:直接计算:则分别计算PdxdyQdydzRdzdx,,,,,,,,,(1) 计算时 Pdxdy,,,
,(?)将曲面投影在面(且只能投影面,即使投影为曲线而非区域,此时xoyxoy
,)为区域D,即根据方程解出:zzxyxyD,,(,),(,),并确定曲Pdxdy,0xyxy,,,
面是朝上还是朝下
1计算下列对坐标的曲面积分
22(1),其中?是柱面被平面及所截下的zdxdyxdydzydzdx,,z,0z,3x1,,y,,,
第一卦限内部分的前侧; z,3
解:(1)计算 zdxdy,,,
?在面投影为0,故 zdxdy,0xoy,,y ,
(2) 计算 xdydzx ,,,
,Dyz:01,03,,,,曲面朝投影为 yozyz
2yzDyz,:01,03,,,,,,,,:1xy故,前侧 ,,yz
2xdydzydydz,,,1故 ,,,,,Dyz
131322,,,,dyydzydydz11 ,,,,0000
12,,31ydy(令 yt,cos),0
,3,22,, 3costdt,04
(3) 计算ydzdx,,,
,曲面朝投影为 xozDxz:01,03,,,,xz
2xzDxz,:01,03,,,,,故,右侧 ,,,:1yx,,xz
2故ydzdxxdzdx,,,1 ,,,,,Dxz
13133,22,,dxxdz1,,,1xdxdz ,,,,00004
故 zdxdyxdydzydzdx,,,,,
333,,,,,,,0= zdxdyxdydzydzdx,,,,,,,,442,,,
1222zxy,,()(2),其中?是抛物面介于平面及之z,0z,2()zxdydzzdxdy,,,,2,
间的部分的下侧;
z
,2
,1
y x
22,D解:法一(直接计算):计算,将投影到面为 ,zdxdyxoyxy,,4xy,,,122,,,,zxyxyD:(),,,朝下,故 ,,xy2
22,11222,,,,drrdr4 ,,,,,zdxdyxydxdy(),,,,,,00Dxy22,
Z=2
122,计算将投影到面为,如图 D()zxdydz,yozzy,yz,,,2
2y ,,,,:2,,xzyxyD,其中,朝前 ,,,,,,,yz112
2,,,,,:2,,xzyxyD,朝后,故 ,,yz2
22()()()zxdydzzxdydz,,,, ,,,,,,,,,12
2222,,,(2)zzydydz,,,(2)zzydydz ,,,,DDyzyz
3222122222,,22dyzydz,,22zydydz ,,zydy2(2)121,,,,,2,yD2,2yyz322
3322242222,,(4)ydy,,(4)ydy(其中令) y,2sin,,,,2033
,442,,,,,16cos4d ,03
2故 ()8zxdydzzdxdy,,,,,,,
12222zxy,,()法二(投影面转换法)因为,D:,朝下,,所以 zx,xy,,4xyx2
2 ()zxdydzzdxdy,,,,,
2 ,,,,[()()]zxxzdxdy,,,
22 ,,,,()zxxzdxdy,,,
11222222,,,,,,,xyxxxydxdy[[(()()]],,42Dxy
11222222,,,,,xyxxxydxdy(()()],,42Dxy
11222222 ,,,,,xyxdxdyxxydxdy(()[()],,,,42DDxyxy
1222 ,,,,xxydxdy0[()],,2Dxy
22,222,,,,()8xydxdydrrdr,, ,,,,00Dxy
1222(其中利用对称性:, xyxdxdy,,(()0,,4Dxy
12222222由于:易知:,即) Dxdxdyydxdy,xdxdyxydxdy,,xy,,4()xy,,,,,,,,2DDDDxyxyxyxy
2把对坐标的曲面积分化为对面积的曲面积分: (1)?:平面被柱面PxyzdydzQxyzdzdxRxyzdxdy(,,)(,,)(,,),,zx,,1,,,
22所截部分的下侧; x1,,y
,12解:曲面在处的法向量为,故: (,,)xyz(1,0,1),,,cos,,,2222(1)0(1),,,,
,120,,故 ,,,,cos,,,cos02222222(1)0(1),,,,,,,,(1)0(1)
PxyzdydzQxyzdzdxRxyzdxdy(,,)(,,)(,,),,,,,
,,,[(,,)cos(,,)cos(,,)cos]PxyzQxyzRxyzdS,,,,,,
2 ,,,[(,,)(,,)]PxyzRxyzdS,,2,
(注意对于非定向曲面可为,或,但对于定向曲面朝(1,0,1),,,,(1,0,1)(1,0,1)zx,,1
下则第三个分量应为负)
22PxyzdydzQxyzdzdxRxyzdxdy(,,)(,,)(,,),,(2)?:抛物面被平面所截y,2yxz,,2,,,
的部分的左侧.
解:曲面在处的法向量为,故: (,,)xyz(4,1,2)xz,
44xx ,,,cos22222,,,,,(4)(1)(2)1164xzxz
,12z,coscos,,,,,故 22221164,,xz1164,,xz
PxyzdydzQxyzdzdxRxyzdxdy(,,)(,,)(,,),,,,,
,,,[(,,)cos(,,)cos(,,)cos]PxyzQxyzRxyzdS,,,,,,
4(,,)(,,)2(,,)xPxyzQxyzzRxyz,, ,dS,,221164,,xz,
22(注意对于非定向曲面可为,或,但(4,1,2)xz,,,,,,(4,1,2)(4,1,2)xzxzyxz,,2
对于定向曲面朝做则第二个分量应为负) 3(计算曲面积分 [(,,)][2(,,)][(,,)]fxyzxdydzfxyzydzdxfxyzzdxdy,,,,,,,,
其中为连续函数,?是平面在第四卦限内的上侧. fxyz(,,)xyz,,,1
z 1
y
,1
x
,1 解:由?是平面在第四卦限内的上侧,故曲面在处的法向量为 xyz,,,1(,,)xyz(1,1,1),
333cos,,cos,,,cos,,故,,,则 333
[(,,)][2(,,)][(,,)]fxyzxdydzfxyzydzdxfxyzzdxdy,,,,,,,,
333 ,,,,,,,{[(,,)][2(,,)]()[(,,)]}fxyzxfxyzyfxyzzdS,,333,
331 ,,,,,()xyzdSdS,,,,332,,
,在面投影区域面积xoy,(其中平面的面积为) cos,
2222zxy,,5( 计算()xydzdxzdxdy,,,?为锥面上满足,y,0,z,1x,0,,,
的那部分曲面的下侧.
解:(采用投影面转换法计算较为简单)
yz,由,有 y22xy,
22 ()xydzdxzdxdy,,,,,
22 ,,,,[()()]xyzzdxdyy,,,
22 ,,,,()yxyzdxdy,,,
2222又?为锥面zxy,,:,,,朝下, Dx,0y,0xy,,1xy2222 ()()xydzdxzdxdyyxyzdxdy,,,,,,,,,,,,
,1222222,,,,,,()yxyxydxdy ,,,drrdr,,(1sin),,,,00Dxy
,,113222 ,,drdrdrdr,,,sin,,,,0000
,1,,132,,,,sindrdr ,,,,00646
?6 Gauss公式与Stokes公式
1(利用高斯公式计算下列曲面积分.
333222(1)其中?是球面的外侧. xdydzydzdxzdxdy,,xyz,,,1,,,
333解 xdydzydzdxzdxdy,,,,,
222,,,(333)xyzdxdydz ,,,,
222,,,3()xyzdxdydz ,,,,
21,,22,ddrrdr,,,sin ,,,000
2223()3xyzdxdydzdxdydz,,,(本题中若写成是错误的,为什么,) ,,,,,,,,
22222zy,,xzy,,,2x2)其中?为由曲面与所围2xzdydzyzdzdxzdxdy,,,,,
立体的表面的外侧.
2解: 2xzdydzyzdzdxzdxdy,,,,z,2 ,
,zdxdydz,,,(22)zzzdxdydz ,,,,,,,,
z,1 (若采用先二后一的方法计算三重积分)
22,其中 ,,,,,:01,:zDxyz,,,,,z112
22 ,,,,,,:12,:2()zDxyzz2
12zdxdydz ,,dzzdxdydzzdxdy,,,,,,,,,,DD01zz
1212,32,,,,zdzzzdz(2) ,,zdzdxdyzdzdxdy,,,,,,,,,,01DD01zz2
(若采用柱坐标方法计算三重积分)
2 ,,,,,,,,:2,01,02rzrr,,
2212,,r, zdxdydzdrdrzdz,,,,,,,,,,00r22(计算下列曲面积分:
222(1),?是球面的上侧. yzdzdxdxdy,2xyzz,,,,4(0),,,
22,,,,:0,:4zDxy解;作曲面,朝下。则 xy1
yzdzdxdxdy,2,,, ,
,,yzdzdxdxdy2,,yzdzdxdxdy2 ,,,,,,,,11
,zdxdydz,,yzdzdxdxdy2 ,,,,,, ,,11
222其中 ,,,,,:4(0)xyzz
222zdxdydz,dzzdxdy,,zzdz,,4(先二后一) ,,,,,,,,D00z
22,,,,:0,:4zDxy由,朝下,有 xy1
yzdzdxdxdy,2,,02dxdy,故 ,,,,28dxdy,,,,,,,,,D11xy
yzdzdxdxdy,,212,,,,
32222(2),?为抛物面被平面所截下的xdydzxzdzdxydxdy,,23z,0zxy,,,4,,,
部分的下侧.
22,,,,:0,:4zDxy解;作曲面,朝上。则 xy1
322 xdydzxzdzdxydxdy,,23,,,
,
322322 ,,,xdydzxzdzdxydxdy23,,,xdydzxzdzdxydxdy23,,,,,,,,11
2322 ,,,,(300)xdxdydz,,,xdydzxzdzdxydxdy23,,,,,,,1
22其中 ,,,,,:4(0)zxyz
2224,,r222 xdxdydzdrdrrdz,,,cos,,,,,,,000
22224,,r16,3223,,,rrdr4 ,cos,,drdrdz,,,,,,,00003
2(用柱坐标) ,,,,,,,,:04,02,02,zrr,,
22,,,,:0,:4zDxy,朝上有 由xy1
3222xdydzxzdzdxydxdy,,23,,,003ydxdy ,,,,,,11
22,22,23232,,3sin12,,,drdr ,,33sinydxdydrdr,,,,,,,,0000Dxy
322故 xdydzxzdzdxydxdy,,23,,,,,161228,,,,,,
(其中利用定积分的几何意义有
222,,,1222,,,,,,,,,,,,cossin(sincos)ddd) ,,,0002
22323:计算曲面积分其中?为和xzdydzxyzdzdxxyyzdxdy,,,,()(2)z,0,,,
222zaxy,,,所围曲面外侧. a
2232解: xzdydzxyzdzdxxyyzdxdy,,,,()(2),, a,
,2a,222222 ,,,,()sinzxydxdydzddrrdr,,,,,,,,,,000 a
a,2a,22252,,ddrrdra,,,,sin ,,,0005
4(设是连续可导函数,计算曲面积分 f
11yy333其中?为锥面 ,,,xdydzfydzdxfzdxdy[()]+[()],,zzyz,
22222222xyz,,与两球面及所围立体表面的外侧. xyz,,,1xyz,,,4
11yy333解: ,,,xdydzfydzdxfzdxdy[()]+[()],,zzyz,
11yy2'2'2,,,,, (3()3()3)xfyfzdxdydz22,,,,zzzz
222 ,,,(333)xyzdxdydz,,,,
,22,224 ,ddrrdr,,,sin,,,001
,22,93(22),224,,,,,,ddrrdrsin ,,,00155(利用斯托克斯公式计算下列曲线积分:
2222,xyza,,,,(1),为圆周:,从z轴正向看去,取逆时针方ydx,zdy,xdz,,,xyz,,,0,向.
dydzdzdxdxdy
,,,解:原积分= ,,,,,,,,dydzdzdxdxdydydzdzdxdxdy,,,,,,,,,xyz,,,
yzx ,,,,:0xyz
,(其中如图它是在球内的部分,朝上。) xyz,,,0
,1,1,1的法向量为,故 ,,
333 ,,,,,,,,dydzdzdxdxdydS(),,,,333,,
2 ,,,,33dSa,,,,
222,xya,,,,(2),为椭圆(,0)ab,,从z轴正向(y,z)dx,(z,x)dy,(x,y)dz,,xz,,,1,ab,看去,取逆时针方向.
dydzdzdxdxdy
,,,222dydzdzdxdxdy 解:原积分=,,,, ,,,,xyz,,,,,
yzzxxy,,,
xz,,,1,,,,2dydzdzdxdxdy(其中它是在圆柱内的部分,朝上) ,,ab,
,ba,0,的法向量为,故 ,,
ba原积分 ,,,,,,,22()dydzdzdxdxdydS,,,,2222abab,,,,
2baa,ba ,,,,,,2()2()aab,,,2()dS,,,22222222aabab,,abab,,,22ab,
第十章 自测题
xtt,cos,
,,ytt,sin1((1)求zds,其中为曲线,; (0),,tt,0, ,,zt,,
t022zds,,,,,tttttttdt(cossin)(sincos)1解: ,, ,0
332222t,2t,,(2)22,,t0t200,,ttdt2 ,,0,033
xx222(2)求(sin2)(cos2)eyydxeydy,,,,其中L为上半圆周,()xaya,,,,L
,沿逆时针方向. y,0
,,QPxx,,2解:, PeyyQey,,,,sin2,cos2,,xy
做直线段,则 OAyxa:0,:02,,
xx(sin2)(cos2)eyydxeydy,,, ,L
xxxx,,,,(sin2)(cos2)eyydxeydy,,,,(sin2)(cos2)eyydxeydy ,,,LOAOA
xx2a ,2dxdy,,,,(sin2)(cos2)eyydxeydy ,,,DOA
2xx,,,,,,aeyydxeydy(sin2)(cos2) ,O A OA
由OAyxa:0,:02,,有
2axxx(sin2)(cos2)eyydxeydy,,,,,,,,(sin020)00edx ,,OA0
xx2(sin2)(cos2)eyydxeydya,,,,,故 ,L
2(计算下列各题:
dS222(1),其中?为界于z,0与zHH,,(0)之间的柱面:. xyR,,,,222,,xyz,
zH,
y
dSdS解:利用对称性有 ,4,,,,222222xyzxyz,,,,,,1
,其中为在第一卦限部分如图 ,1
22 ,,,,,,,:,:0,0xRyDyRzH1
2,,,yR222,,,,,10dydz dSxxdydz,,,1,dydzyz2222,,Ry,Ry,,,dSdSdS ,,44,,,,,,22222222xyzxyzRz,,,,,,,,11
HR1R1R ,4dydz,4dzdy,,,,2222222200RzRy,,RzRy,,D
HR1R ,4dzdy,,222200RzRy,,
RHy,,22 4lnarcsin,,,zRzR,,,,0R,,0
22HRH,, 2ln,R,R
222(2)求其中?为锥面()()()yzdydzzxdzdxxydxdy,,,,,,,,
22zxyzh,,,,(0)的外侧. ,,:zh1
222解:作曲面,朝上,则 ,,,,:,:zhDxyh1
222 ()()()yzdydzzxdzdxxydxdy,,,,,,,22,,,:zxy ,
222,,,,,,()()()yzdydzzxdzdxxydxdy ,,,,,1
222,,,,,,()()()yzdydzzxdzdxxydxdy ,,,1
222,,,(000)dxdydz,,,,,,()()()yzdydzzxdzdxxydxdy ,,,,,,,1
222,,,,:,:zhDxyh由,朝上有 1
222()()()yzdydzzxdzdxxydxdy,,,,, ,,,1
2 ,,,,00()xydxdy,,,1
222 ,,()xydxdy,,,,xdxdyydxdyxdxdy0,,,,,,,,DDDD
4,h211,h223 ,,,,()xydxdydrdr,,,,,00224D
4h,222故 ()()()yzdydzzxdzdxxydxdy,,,,,,,,,4,
22xdydzydzdxzdxdyzxy(2)(1),,,,(3)其中?为曲面的上侧. 1(0),,,,z,,22235169()xyz,,,
22(2)(1)xy,,解:作曲面,,,,:0,:1zD,朝下,则 1169xdydzydzdxzdxdy,, ,,2223()xyz,,,
xdydzydzdxzdxdyxdydzydzdxzdxdy,,,, ,,,,,,22232223()()xyzxyz,,,,,,,,11
xdydzydzdxzdxdy,, 0dxdydz,,,,,,,2223,()xyz,,,1
22(2)(1)xy,,,,,,:0,:1zD由,朝下,有 1169
xdydzydzdxzdxdy,, ,,2223()xyz,,,1
0dxdy ,,,,000,,2223,,()xyz,1
xdydzydzdxzdxdy,,故=0 ,,2223()xyz,,,
222zaxy,,,3(求均匀曲面的质心的坐标.
zdS,,,222222,,,,,,,,:,:zaxyDxya解:利用对称性,z,其中 xy,,0,0
dS,,,,
a22 1dSzzdxdydxdy,,,,xy222axy,,
zdS,,,, ,z
dS,,,,
a222,axydxdy,,,,222Daxy,, ,adxdy,,,222Daxy,,
dxdy,,D ,1dxdy,,222Daxy,,
2,a ,,2arddr,,,2200,ar
2a,aa(0,0,),故质心为 ,,a22222(),,ar,0