【正文】
f(? ? ????? ???101023dy)dxxgyx f(? ? ?????????????101022221022dy)dxxgyf|xgyf(?? ? ? ?? ???? ??D1010224224dx)y,x(gyxfdyd xd y)y,x(gyxfI? ? ??????10102222dy)dxxgyf( ????????D2222d x d yxgyf(分部積分) (分部積分) ,)y1(y2x g22?????又由分部積分法得 ? ? ?????102210dx]dyyf))y1(y2([I? ? ????????? ??10101y0y dx]dyyf)y21(|yf)y1(y[2? ? ??????1010dx]dyyf)y21([2? ??1010dx)dy)y,x(f(4 ???Dd x d y)y,x(f4故 ???? ???? ??D224D|d xd y)y1)(x1(xyyxf|41|d xd y)y,x(f|?? ???????D224d xd y|)y1)(x1(xy||yxf|41?? ????Dd x d y)y1)(x1(xyB41? ??102]dx)x1(x[4B .1 4 4B?21,dy)xy(u)y(u1)x(u,]1,0[)x( 1x?????? ? 試證且連續(xù)在設(shè)例證 ??10dx)x(ua 令? ? ???101dx])x(u)([1a dyyyux? dyyyuxx)x(u)(d1 110???? ???則令 ,xyt ??? ? ????100yd t ))(t(udy)y(u1a ? ????10y0dt)t(udy)y(u1 102y0|)dt)t(u(211 ????? 2a21 ???有實(shí)數(shù)解于是一元二次方程 01xx2 2 ????,]1,0[)x(u 連續(xù)在由于 ?10dx)x(u 存在,故0241 ??????.21??即??????????222 Ryx222R 0)yxyx(x d yy d xlim14 證明例證 ,)(,)( 222222 yxyx xQyxyx yP ?? ?????令222222 Ryx22222Ryx22 |)(yx|QP???? ?????yxyx22 )xyR(R?? 22 )|xy|R(R??2222 )2yxR(R?? 3R4?dsQP)( 22222222222 ???????????RyxRyxyxyxxdyy d x23 R8R2R4 ?? ????????????222 Ryx222R 0)yxyx(x d yy d xlim所以,0R8l i m2R ?????而? ?年考研題))(的正向邊界,試證:為已知平面區(qū)域例2022.(2dxyedyxe)2(。dxyedyxedxyedyxe1DL,y0,x0)y,x(D152Lxs i nys i nLxs i nys i nLxs i nys i n???????????????????解 由格林公式,得)1(??? ?? ???DxyLxy dxdyeedxyedyxe )( s i ns i ns i ns i n??? ??? ??DxyLxy dxdyeedxyedyxe )( s i ns i ns i ns i n?? ??Dxy dxdyee )( s i ns i n ?? ?? ?Dxy d x d yee )( s i ns i n對(duì)稱,所以關(guān)于因?yàn)閰^(qū)域 xyD ?.s i ns i ns i ns i n ?? ??? ??LxyLxy dxyedyxedxyedyxe( 2) 由( 1)知 ? ??L xy dxyedyxe s i ns i n ?? ???Dxy d x d yee )( s i ns i n???Ddxdy2?? ???Dxx dxdyee )( s i ns i n .22??12222 ?? byax2 0 1 1uu 39。39。yy39。39。xx ?? ???L dsnu nu?? 設(shè) u在閉曲線 L: 所圍閉區(qū)域 D上有連續(xù)二階偏導(dǎo)數(shù) , 且 , 求 . 其中 為 u沿 D邊界外法線的方向?qū)?shù) . 例 16 )y,x(P }s in,c o s{T ???}s in,c o s{n ??? 2??? ??dsLs i nyuc osxudsL nu ?????????????????? ??解:設(shè)曲線 L上任意一點(diǎn) 處的切向量為 法向量為 ,注意到 ,于是 dsLc osyus i nxu? ?????????????? ??? ??????????????Ldxyudyxu??? ?????????? ????????D2222Lba2022d x d yy ux udsnu再由 Green 公式有 .