20xx大學數(shù)學習題一答案精選(編輯修改稿)
2025-03-25 22:11 本頁面
【文章內(nèi)容簡介】 4?dx??11?560?x3?x5?35????015(2)如圖111所示，L=L1+L2．其中L1的參數(shù)方程為圖111?x?a?acost?0?t?π?y?asintL2的方程為y=0(0≤x≤2a)故 ??Lxydx??Lxydx?1?Lxydx2??π0a?1+cost?asint??a?acost??dt??2a00dx??π0a3??sin2t??1?cost?dt??a3??π2sintdt??π2sintdsint???π2a3π(3)?ydx?xdy?2L?0??Rsint??Rsint??RcostRcost??dtπ?R2?20cos2tdtπ?R2?1?2?sin2t?2??0?0(4)圓周的參數(shù)方程為：x=acost，y=asint，t:0→2π． 故 ???x?y?dx??x?y?dyLx2?y2?1a2?2π0???acost?asint???asint???acost?asint?acost??dt?12π2a2?0??a?dt??2π262(5)?2?xdx?zdy?ydz??π220?k??k?asin??a??sin???acos?acos??d???π322?k??a?d???1π3?3?a2???k?3??0?133kπ3?a2π?x?3t(6)直線Γ的參數(shù)方程是??y?2t t從1→0．??z?t故?322?xdx?3zydy???xy?dz??0?1?27t3?3?3t?4t2?2???9t2?2t???dt??87t31dt10?87?4t41??874(7)??AB?BC?CA(如圖112所示)圖112AB:?y?1?x??z?0，x從0→1 ?ABdx?dy?ydz??1??1???1???dx??2．BC:?x?0?，z從0→1 ?y?1?z 263?dx?dy?ydz?1?zBC?1??0???1??????dz??1?2?z?dz??112??2z?z?2??0?32CA:?y?0?，?z?1?xx從0→1 ?1CAdx?dy?ydz???1?0?0?dx?1．故??dx?dy?ydzL????dx?dy?ydzAB??BC?CA????2??312?1?2(8)??x2?2xydx?2xyL??y?2?dy??1?2?2x?x222x??1??x???x4?2x?x???dx??1?x2?2x3?2x5?4x4?1?dx??14154．計算??x?y?dx??y?x?dy，其中L是L(1)拋物線y2=x上從點(1,1)到點(4,2)的一段?。?2)從點(1,1)到點(4,2)的直線段；(3)先沿直線從(1,1)到點(1,2)，然后再沿直線到點(4,2)的折線；(4)曲線x=2t2+t+1,y=t2+1上從點(1,1)到點(4,2)的一段弧．解：(1)L：??x?y2，y：1→2，故?y?y?L?x?y?dx??y?x?dy??2?y21???y??2y??y?y2??1??dy??23?y2?y?dy1?2y2??141312??2y?y?y?32??1?343(2)從(1,1)到(4,2)的直線段方程為x=3y2，y：1→2264故??x?y?dx??y?x?dyL????3y?2?y??3??y?3y?2???dy??12??10y?4?dy1222????5y?4y?1?11(3)設從點(1,1) 到點(1,2)的線段為L1，從點(1,2)到(4,2)的線段為L2，那么L=L1+L2．且L1：??x?1?y?y，y：1→2；L2：??x?x?y?2，x：1→4；故??x?y?dx??y?x?dyL1????1?y??0??y?1???dy??12?21?y2??y?1?dy???y??2?12?12??x
范文總結(jié)相關推薦
鄂ICP備17016276號-1