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1曲线积分与曲面积分计算下列曲线积分.1.0,:,)(.13234323434yayxLdsxyxL且解:.xdsxx,yLL0奇函数是轴对称关于dsyxdsxyxLL)()(34343434;ttsinaytcosax:L033dtttattadttytxds222222)cossin3()sincos3()()(dttcostsina3dtttatatadsyxLcossin3)sincos()(04434343434dtttttacossin)sin(cos304437tdttttacossin)sin(cos6237044.adttcostsinadttsintcosa3737372660505.0)()(:)(.2222222xyxayxLdsyxL:解.0,LydsyyxL是奇函数关于轴对称关于LLxdsdsyx)(2cosarsin,cosryrxcossin)(,sincos)(rryrrx22)sincos()(rrx2222sinsincos2cosrrrr22)cossin()(rry2222cossincos2sinrrrrdyxds)()(22drr222cos)2cos()2cos2sin(22222aaarrdads2cosdaaxdsL2coscos2cos44.2222cos22244aada..2;.1:,,0,2,.32222轴的转动惯量关于构件的质量构件求且线密度所围区域的边界曲线为由设xLLyxyaxyxxyL解:LdsyxmL22.1的质量构件dsyxLLL22)(1232202221axdxdsyxaL22222LLdsaxdsyxdttatataaa220)cos()sin(cos22dtta202cos12.22sin4cos2cos2220220220222222aadtadtattt20022221123axdxdxxdsyxaaL.)223(2amdsyxyIxLLx222.2轴的转动惯量关于构件dsyxyLLL222)(12301222Ldsyxy2222222LLdsaxydsyxydttatataaata2202)cos()sin()cos(2)sin(2dttadttatt202422024cossin2cos2sin222dtattt202224cos)cossin2(222222024sincossin162tttda.1522sin)sin1(sin16422220242adattt.21222403302223adxxdxxdsyxyaaL.)152221(4aIx.)0,0(,.4222222222逆时针方向轴正向向负向看去为交线从与是其中求xazaxyxazyxdzxdyzdxy解:20:sincos22222ttytxaxyxaaa22222222222)sin()cos(azttazyxaaa代入22422224)cos22(]sin)cos1[(22aztazttaa2222222sinsin2cos1ttazaztaz20:sinsincos:2222taztytxtaaadzxdyzdxy222dtattatttaaataaa])sin()cos()sin()sin()cos()sin[(2222220222222dtttatttaaaataa)]cos()cos()cos()sin()sin)(sin[(22222220222222.4]coscossinsin[323212022213413adtttatt20:sinsincos:2222taztytxtaaa.,,,.)4,3()2,1(),(.52所做的功对质点求力角小于轴的正向夹其与与原点的连线其方向垂直于质点的距离与原点的大小等于质点作用的过程中受变力运动从如图红线所示为直径的半圆周沿以设质点PFyPPFFBAABP:解.443:sin23cos22:yxLAB为直径的半圆周).,(,),(xyFLyxPLxdyydxW443])sin23)(cos22()cos22)(sin23([d).1(2)]cos2)(cos22()sin2)(sin23([443d.)0,()0,0(sin:,))ln((.1.62222的一段至上由其中计算下列曲线积分xyLdyyxxxyydxyxIL:解,利用格林公式dyyxxxyydxyxIAOLAO))ln(()(2222;)(2222yxyyxy)1()))ln(((222222yxxyxxyyyxxxyyx.222yxyyAOLdyyxxxyydxyx))ln((2222Ddxdyyxyyxyy)(22222Ddxdyy2Ddxdyy2AOLdyyxxxyydxyx))ln((2222dyydxxsin0200203cos)cos1(31sin31xdxxdx.94.2))ln((202222xdxdyyxxxyydxyxAO)2(94))ln((22222Ldyyxxxyydxyx.9422.)0,0(1)()1()2,2(,]cos[sin,.2222一段的的左半圆至圆周沿是从计算具有一阶连续导数设OyxALdyxyxfydxxfxfL:解yxfxyxfxQxcos]cos[yxfyxfyPycos]sin[由格林公式LBAOBLdyxyxfydxxf]cos[sindxdyyPxQD)()221()(2DDdxdydxdydyxyxfydxxfIOBBA]cos[sin)()221(2,00]cos[sin20dxdyxyxfydxxfOB,00]cos[sin20dxdyxyxfydxxfOB20]2cos2[]cos[sindyyfdyxyxfydxxfBA.4]2sin2[220yyf.21234)221(4222I.)3,1()43,4(,)1()2(1,),(.7222的直线段至从是其中计算上有连续导数在设BALdyxyfyyxdxxyfyyIxfL解:;1))1((222xyfxyxyfyxyfyyxxQx.2))2(1(22xyfxyxyfyxyfyyyPy,1:1LdxyI令dyxyfyyxdxxyfyyIL)1()1(1222221III14:,3::,12xxyLI取路径与路径无关dyxyfyyxdxxyfyyIL)1()1(122221dyyxxyxfdxxyyfyL)()1(21dyyxxyxfdxxyyfyL)()1(21dxxxxxffxxI]3)33(333[2214214:3:1xxxxyLdxxfxfxx]333333[14.53214dxxLdxyI11343:345:yyyyxL.2ln3834113143dyydxyIL.2ln385I.,.,,:,,,,.822,20,02,0,02yxQdyyxQdxxyedyyxQdxxyeRtRdyyxQdxxyexoyyxQtxtxLx求恒有且内与路径无关在曲线积分导数平面上具有一阶连续偏在设:解xxeyPxQR所以内曲线积分与路径无关因为在,2yxedxxedxxQyxQxx)1(,故2,0,02,txdyyxQdxxye由题中等式左2020,02dyytQdxextx202])1([2dyytet202)1(22dyytet2,20,0,txdyyxQdxxye由题中等式右2020,20txdyyQdxex202])12([tdyye2022tdyyte22022202)1(2ttdyytedyyte,2ut令uudyyuedyyue0220)1(2即ueueuu22:求导得两边关于22eyeyy.2)1(,2eyexeyxQyx故.,52.1.922原点简单闭曲线为逆时针方向的不过其中计算LyxydxxdyIL:解;522552.12222222yxxyyxxxxQ;5225522222222yxxyyxyxyPyPxQ:,由格林公式得不包含原点时当L.05225522552222222222222dxdyyxxyyxxyyxydxxdyLDLxyoLLD.)5,2()5,4(32:,52.2222的一段至点上从点其中计算BAxxyLyxydxxdyIL:,由格林公式得包含原点时当L.05225522552222222222222dxdyyxxyyxxyyxydxxdyLLDLL.02:5sin2cos:ttytxLLLyxydxxdyyxydxxdy22225252dttttt022222sin5sin5cos2cos.102Lxyo2525LLLD.)5,2()5,4(32:,52.2222的一段至点上从点其中计算BAxxyLyxydxxdyIL10252.222BALyxydxxdy:解BALyxydxxdyyxydxxdy22225210252422255250102xdx42212525102xdx42552arctan551125102x).5524arctan5522(arctan101102.),(),(:,),(),,(,,),(),(max.1022,MLdyyxQd
本文标题:第八章、曲线积分与曲面积分 Microsoft PowerPoint 演示文稿
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