求曲面积分I=∫∫1/yf(x/y)dyzy+1/xf(x/y)dzdx+zdxdy，如图 计算曲面积分 I=∫∫(S+) (x)dydz+(y+y^2...
\u6c42\u66f2\u9762\u79ef\u5206I=\u222b\u222b { (xdydz+ydzdx+zdxdy) / \u221a\uff08x²+y²+z²\uff09},\u5176\u4e2d\u4e3a\u4e86\u5229\u7528\u9ad8\u65af\u516c\u5f0f\uff0c\u5c06\u76ee\u6807\u66f2\u9762\u8865\u6210\u5c01\u95ed\u7684\u66f2\u9762\uff0c\u4e14\u65b9\u5411\u5411\u5916\u4fa7\uff0c\u6700\u540e\u79ef\u5206\u503c\u51cf\u53bb\u8fd9\u4e00\u90e8\u5206\u5373\u53ef.
\u76ee\u6807\u66f2\u9762\u4e3a\u534a\u7403\u9762\uff0c\u8865\u5145\u534a\u7403\u9762\u7684\u5e95\u9762\u90e8\u5206\uff0c\u8bbe\u4e3a\u2211a. \u65b0\u5f62\u6210\u7684\u5c01\u95ed\u66f2\u9762\u8bbe\u4e3a \u2211b. \u5728\u5e95\u9762\u65f6\uff0cz = 0\uff0cdz = 0.
\u5219\uff1a\u539f\u79ef\u5206 I = \u222b\u222b(\u2211b)xdydz+ydzdx+zdxdy - \u222b\u222b(\u2211a)xdydz+ydzdx+zdxdy
= \u222b\u222b\u222b 3 dV - 0
= 3V(\u534a\u7403)
= 2\u03c0R^3.
\u5e0c\u671b\u80fd\u5e2e\u5230\u4f60
\u7528\u9ad8\u65af\u516c\u5f0f\u505a\uff0c\u8865\u5145\u5e73\u9762S\u2018\uff1az=1\uff0c\u53d6\u4e0a\u4fa7\uff0c\u5219S+S\u2019\u6784\u6210\u4e00\u95ed\u66f2\u9762\uff0c\u5bf9\u5176\u4f7f\u7528\u516c\u5f0f\u516c\u5f0f\uff0c\u4ee4P=x\uff0cQ=y+y^2\uff0cR=z\uff0c\u5219ðP/ðx+ðQ/ðy+ðR/ðz=3+2y\uff0c\u6240\u4ee5I+i'=\u222b\u222b\u222b(3+2y)dxdydz=3\u222b\u222b\u222bdxdydz+\u222b\u222b\u222b2ydxdydz\uff0c\u7b2c\u4e00\u4e2a\u79ef\u5206=3\u222bdz\u222b\u222bdxdy=3\u222b\u03c0zdz=3\u03c0/2\uff0c\u7b2c\u4e8c\u4e2a\u79ef\u5206\u88ab\u79ef\u51fd\u6570\u662f\u5173\u4e8ey\u7684\u5947\u51fd\u6570\uff0c\u79ef\u5206\u533a\u57df\u5173\u4e8exoz\u5e73\u9762\u5bf9\u79f0\uff0c\u6545\u79ef\u5206=0\uff0c\u6240\u4ee5I+i'=3\u03c0/2\u3002\u518d\u8ba1\u7b97l'\uff0c\u5728z=1\u5e73\u9762\u4e0adz=0\uff0c\u6240\u4ee5l'=\u222b\u222bdxdy=\u03c0\uff0c\u6240\u4ee5l=3\u03c0/2-\u03c0=\u03c0/2
积分区域是与y轴同轴的两个圆锥体所截的体积。所求截面可联立两方程解得：y = 8-x^2-z^2 = 6+x^2+z^2
x^2+z^2 = 1
因为截面是一个圆心在原点的圆，可设
x = r cos t
z = r sin t
从此圆的方程可见： r from 0 to 1, t from 0 to 2pi
绛旓細绉垎鍖哄煙鏄笌y杞村悓杞寸殑涓や釜鍦嗛敟浣撴墍鎴殑浣撶Н銆傛墍姹傛埅闈㈠彲鑱旂珛涓ゆ柟绋嬭В寰楋細y = 8-x^2-z^2 = 6+x^2+z^2 x^2+z^2 = 1 鍥犱负鎴潰鏄涓涓渾蹇冨湪鍘熺偣鐨勫渾锛屽彲璁 x = r cos t z = r sin t 浠庢鍦嗙殑鏂圭▼鍙锛 r from 0 to 1, t from 0 to 2pi ...
绛旓細绠鍗璁＄畻涓涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細鎴戠殑 楂樻暟 鏇查潰绉垎1 鎴戞潵绛 1涓洖绛 #璇濋# 鎵撳伐浜哄繀鐪嬬殑鑱屽満銆庣淮鏉冦忔寚鍗!鐒￠ 2016-06-25 路 TA鑾峰緱瓒呰繃296涓禐 鐭ラ亾灏忔湁寤烘爲绛斾富 鍥炵瓟閲:581 閲囩撼鐜:100% 甯姪鐨勪汉:507涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 鏇村杩介棶杩界瓟 杩介棶 绛旀鏄 鏈洖绛旂敱鎻愰棶鑰呮帹鑽 宸茶禐杩 宸茶俯杩...
绛旓細璁＄畻鏇查潰绉垎I=鈭埆D(x+|y|)dS,鍏朵腑鏇查潰D:|x|+|y|+|z|=1 璁＄畻鏇查潰绉垎I=鈭埆D(x+|y|)dS,鍏朵腑鏇查潰D:|x|+|y|+|z|=1鍦ㄧ嚎绛,姹傝В绛!... 璁＄畻鏇查潰绉垎I=鈭埆D(x+|y|)dS,鍏朵腑鏇查潰D:|x|+|y|+|z|=1 鍦ㄧ嚎绛,姹傝В绛! 灞曞紑 1涓洖绛 #鐑# 缇庝緷绀艰娊浜烘皵鏂眰绗1,濡備綍璇勪环濂圭殑琛ㄧ幇?涓...
绛旓細楂樻柉鍏紡璁＄畻鏇查潰绉垎I=鈭埆-ydxdz+(z+1)dxdy 鍏朵腑危鏄渾鏌遍潰 x^2+y^2=4 琚玿+z=2鍜寊=0鎵鎴嚭閮ㄥ垎鐨勫渚ч椤 鍦ㄩ棶 鍏ㄩ儴闂 濞变箰浼戦棽 娓告垙 鏃呮父 鏁欒偛鍩硅 閲戣瀺璐㈢粡 鍖荤枟鍋ュ悍 绉戞妧 瀹剁數鏁扮爜 鏀跨瓥娉曡 鏂囧寲鍘嗗彶 鏃跺皻缇庡 鎯呮劅蹇冪悊 姹借溅 鐢熸椿 鑱屼笟 姣嶅┐ 涓夊啘 浜掕仈缃 鐢熶骇...
绛旓細鍏朵粬1鏉″洖绛 2013-06-18 09:30 miao_8866 | 鍏骇 S鐨勮竟鐣鏇查潰涓篠1:z=(x^2+y^2)^0.5, 0<=z<=1 S2:z=1 ,x^2+y^2銆1 S1鍜孲2鍦▁oy骞抽潰涓婄殑鎶曞奖涓:x^2+y^2銆1 鐢卞绉版:鈭埆xdS=0 杩介棶 浜,鎴戦棶鐨勬槸绗竴棰,璨屼技绛旀鏈夌偣涓嶅搴斿摝,涓嶈繃璋㈣阿浣犱簡銆 璇勮 | 鏇查潰...
绛旓細鎮ㄥソ锛岀瓟妗堝鍥炬墍绀猴細寰堥珮鍏磋兘鍥炵瓟鎮ㄧ殑鎻愰棶锛屾偍涓嶇敤娣诲姞浠讳綍璐㈠瘜锛屽彧瑕佸強鏃堕噰绾冲氨鏄鎴戜滑鏈濂界殑鍥炴姤銆傝嫢鎻愰棶浜鸿繕鏈変换浣曚笉鎳傜殑鍦版柟鍙殢鏃惰拷闂紝鎴戜細灏介噺瑙ｇ瓟锛岀鎮ㄥ涓氳繘姝ワ紝璋㈣阿銆傗槅鈱抇鈱掆槅 濡傛灉闂瑙ｅ喅鍚庯紝璇风偣鍑讳笅闈㈢殑鈥滈変负婊℃剰绛旀鈥
绛旓細琛ュ厖鏇查潰鈭1锛歺2+y24锛1z锛0锛屽彇涓嬩晶锛屽垯锛I=?鈭+鈭1xzdydz+2zydzdx+3xydxdy-?鈭1xzdydz+2zydzdx+3xydxdy=?惟(z+3z)dxdydz+?D3xydxdy锛屽叾涓紝惟 涓衡垜涓庘垜1鎵鍥存垚鐨勭┖闂村尯鍩燂紝D={锛坸锛寉锛墊x2+y24鈮1}涓衡垜1鍦▁Oy闈笂鐨勬姇褰憋紝鍥犱负D鍏充簬x杞村绉帮紝3xy鍏充簬x涓哄鍑芥暟锛屾墍浠ワ細?D3xydxdy...
绛旓細绗竴绫鏇查潰绉垎涓紝琚Н鍑芥暟涓1鐨勬椂鍊欙紝绉垎缁撴灉灏辨槸鏇查潰闈㈢Н銆傚悓鐞嗭紝绗竴绫绘洸绾跨Н鍒嗕腑锛岃绉嚱鏁颁负1灏辨槸鏇茬嚎闀垮害銆傞亾鐞嗗緢绠鍗曪紝鍥犱负寮ч暱鍙互鐞嗚В鎴愬綋绾垮瘑搴︿负1鏃剁殑鏇茬嚎璐ㄩ噺锛岃岄潰绉湪鏁板间笂灏辨槸闈㈠瘑搴︿负1鏃剁殑鏇查潰璐ㄩ噺銆傚綋绾垮瘑搴︽垨闈㈠瘑搴︿笉鏄1鑰屾槸鍑芥暟鏃讹紝绗竴鍨嬬Н鍒嗙粨鏋滃氨鏄嚎鎴栭潰鐨勮川閲忋傜浜岀被...
绛旓細瑙ｏ細鍘熷紡=鈭埆鈭<鈭>(1+1+1)dxdydz (搴旂敤濂ラ珮鍏紡)=3鈭<0,2蟺>d胃鈭<0,1>rdr鈭<r^2,1>dz (浣滄煴闈㈠潗鏍囧彉鎹)=6蟺鈭<0,1>(1-r^2)rdr =6蟺(1/2-1/4)=3蟺/2銆
