曲線積分的計算方法如何來記憶

05-20

曲線積分的計算方法如何來記憶

來自專欄 fulyMath

聲明：本文為原創文章，首發於微信公眾號「湖心亭記」

篇文章還是為了回答一個學生問的問題。就是曲線積分公式那麼複雜，情況多種，怎麼去記住他們呢？今天我就來好好講一下這個問題。相信一直記不住曲線積分計算公式得同學看了一定會有驚喜收穫的。好了，廢話不多說，老師就把自己的幹活無私分享給有需要的同學吧。用心做數學教育，我一直在堅持。

===============

首先我把教材上的曲線積分公式給原原本本的貼出來。大家看下情況還是挺多的。如下：

1、對弧長的曲線積分（第一類）

（1）如果L由y=y(x)給出，x屬於[a,b]

$% MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqi-v0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakeaadaWdrbqaaiaadAgadaqadaqaaiaadIha % caGGSaGaamyEaaGaayjkaiaawMcaaiaadsgacaWGZbaaleaacaWGmb % aabeqdcqGHRiI8aOGaaGjcVlaayIW7caaMi8UaaGjcVlaayIW7cqGH % 9aqpdaWdXaqaaiaadAgadaqadaqaaiaadIhacaGGSaGaamyEamaabm % aabaGaamiEaaGaayjkaiaawMcaaaGaayjkaiaawMcaaaWcbaGaamyy % aaqaaiaadkgaa0Gaey4kIipakmaakaaabaGaaGymaiabgUcaRmaabm % aabaGabmyEayaafaWaaeWaaeaacaWG4baacaGLOaGaayzkaaaacaGL % OaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaqabaGccaWGKbGaamiEaa % aa!67B4! intlimits_L {fleft( {x,y} ight)ds} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} = int_a^b {fleft( {x,yleft( x ight)} ight)} sqrt {1 + {{left( {yleft( x ight)} ight)}^2}} dx$

（2）如果L由x=x(y)給出，y屬於[c,d]，

$% MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqi-v0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakeaadaWdrbqaaiaadAgadaqadaqaaiaadIha % caGGSaGaamyEaaGaayjkaiaawMcaaiaadsgacaWGZbaaleaacaWGmb % aabeqdcqGHRiI8aOGaaGjcVlaayIW7cqGH9aqpdaWdXaqaaiaadAga % daqadaqaaiaadIhadaqadaqaaiaadMhaaiaawIcacaGLPaaacaGGSa % GaamyEaaGaayjkaiaawMcaaaWcbaGaam4yaaqaaiaadsgaa0Gaey4k % IipakmaakaaabaGaaGymaiabgUcaRmaabmaabaGabmiEayaafaWaae % WaaeaacaWG5baacaGLOaGaayzkaaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaaqabaGccaWGKbGaamyEaaaa!6307! intlimits_L {fleft( {x,y} ight)ds} {kern 1pt} {kern 1pt} = int_c^d {fleft( {xleft( y ight),y} ight)} sqrt {1 + {{left( {xleft( y ight)} ight)}^2}} dy$

（3）如果L由 $% MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqi-v0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakeaadaGabaabaeqabaGaamiEaiabg2da9iaa % dIhadaqadaqaaiaadshaaiaawIcacaGLPaaaaeaacaWG5bGaeyypa0 % JaamyEamaabmaabaGaamiDaaGaayjkaiaawMcaaaaacaGL7baaaaa!4B97! left{ egin{array}{l} x = xleft( t ight)\ y = yleft( t ight) end{array} ight.$ ， $% MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqi-v0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakeaacaWG0bGaeyicI48aamWaaeaacqaHXoqy % caGGSaGaeqOSdigacaGLBbGaayzxaaaaaa!47CF! t in left[ {alpha ,eta } ight]$

$% MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqi-v0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakeaadaWdrbqaaiaadAgadaqadaqaaiaadIha % caGGSaGaamyEaaGaayjkaiaawMcaaiaadsgacaWGZbaaleaacaWGmb % aabeqdcqGHRiI8aOGaaGjcVlaayIW7caaMi8UaaGjcVlaayIW7caaM % i8UaaGjcVlabg2da9maapedabaGaamOzamaabmaabaGaamiEamaabm % aabaGaamiDaaGaayjkaiaawMcaaiaacYcacaWG5bWaaeWaaeaacaWG % 0baacaGLOaGaayzkaaaacaGLOaGaayzkaaaaleaacqaHXoqyaeaacq % aHYoGya0Gaey4kIipakmaakaaabaWaaeWaaeaaceWG4bGbauaadaqa % daqaaiaadshaaiaawIcacaGLPaaaaiaawIcacaGLPaaadaahaaWcbe % qaaiaaikdaaaGccqGHRaWkdaqadaqaaiqadMhagaqbamaabmaabaGa % amiDaaGaayjkaiaawMcaaaGaayjkaiaawMcaamaaCaaaleqabaGaaG % OmaaaaaeqaaOGaamizaiaadshaaaa!740B! intlimits_L {fleft( {x,y} ight)ds} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} = int_alpha ^eta {fleft( {xleft( t ight),yleft( t ight)} ight)} sqrt {{{left( {xleft( t ight)} ight)}^2} + {{left( {yleft( t ight)} ight)}^2}} dt$

2、對坐標的曲線積分（第二類）

（1）如果L由y=y(x)給出，x屬於[a,b]

$% MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqi-v0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakqaabeqaamaapefabaGaamiuamaabmaabaGa % amiEaiaacYcacaWG5baacaGLOaGaayzkaaGaamizaiaadIhacqGHRa % WkcaWGrbWaaeWaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPaaa % caWGKbGaamyEaaWcbaGaamitaaqab0Gaey4kIipakiaayIW7caaMi8 % oabaGaaeypamaapedabaWaamWaaeaacaWGqbWaaeWaaeaacaWG4bGa % aiilaiaadMhadaqadaqaaiaadIhaaiaawIcacaGLPaaaaiaawIcaca % GLPaaacqGHRaWkcaWGrbWaaeWaaeaacaWG4bGaaiilaiaadMhacaGG % OaGaamiEaiaacMcaaiaawIcacaGLPaaacqGHIaYTceWG5bGbauaaai % aawUfacaGLDbaaaSqaaiaadggaaeaacaWGIbaaniabgUIiYdGccaWG % KbGaamiEaaaaaa!6F78! egin{array}{l} intlimits_L {Pleft( {x,y} ight)dx + Qleft( {x,y} ight)dy} {kern 1pt} {kern 1pt} \ { m{ = }}int_a^b {left[ {Pleft( {x,yleft( x ight)} ight) + Qleft( {x,y(x)} ight) ullet y} ight]} dx end{array}$

（2）如果L由x=x(y)給出，y屬於[c,d]，

$% MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqi-v0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakqaabeqaamaapefabaGaamiuamaabmaabaGa % amiEaiaacYcacaWG5baacaGLOaGaayzkaaGaamizaiaadIhacqGHRa % WkcaWGrbWaaeWaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPaaa % caWGKbGaamyEaaWcbaGaamitaaqab0Gaey4kIipakiaayIW7caaMi8 % oabaGaaeypamaapedabaWaamWaaeaacaWGqbWaaeWaaeaacaWG4bWa % aeWaaeaacaWG5baacaGLOaGaayzkaaGaaiilaiaadMhaaiaawIcaca % GLPaaacqGHIaYTceWG4bGbauaadaqadaqaaiaadMhaaiaawIcacaGL % PaaacqGHRaWkcaWGrbWaaeWaaeaacaWG4bGaaiilaiaadMhacaGGOa % GaamiEaiaacMcaaiaawIcacaGLPaaaaiaawUfacaGLDbaaaSqaaiaa % dogaaeaacaWGKbaaniabgUIiYdGccaWGKbGaamyEaaaaaa!7204! egin{array}{l} intlimits_L {Pleft( {x,y} ight)dx + Qleft( {x,y} ight)dy} {kern 1pt} {kern 1pt} \ { m{ = }}int_c^d {left[ {Pleft( {xleft( y ight),y} ight) ullet xleft( y ight) + Qleft( {x,y(x)} ight)} ight]} dy end{array}$

（3）如果L由 $[left{ egin{array}{l} x = xleft( t ight)\ y = yleft( t ight) end{array} ight.]$ ， $[t in left[ {alpha ,eta } ight]]$

$[egin{array}{l} intlimits_L {Pleft( {x,y} ight)dx + Qleft( {x,y} ight)dy} {kern 1pt} {kern 1pt} \ { m{ = }}int_alpha ^eta {left[ {Pleft( {xleft( t ight),yleft( t ight)} ight) ullet xleft( t ight) + Qleft( {xleft( t ight),y(t)} ight) ullet yleft( t ight)} ight]} dt end{array}]$

===============

好了，只是貼個公式，就佔用了那麼多篇幅，看來計算公式真的夠冗長的。其實大家仔細觀察上面的公式，無論第一型曲線積分還是第二型曲線積分，都只需要記住第三種情況就行了，因為前兩種都是第三種的特殊形式。那麼這就是今天我要介紹的簡單方法？？哈哈，當然不是，我要介紹的比這個還簡單。

===============

好了，不說廢話。直接上乾貨。

一、第一類曲線積分的計算記憶方法

只需要記住這個式子：

$[intlimits_L {fleft( {x,y} ight)ds} = intlimits_L {fleft( {x,y} ight)} sqrt {{{left( {dx} ight)}^2} + {{left( {dy} ight)}^2}} ]$

關於這個式子是怎麼推導出來的，限於篇幅就不講了，基礎差的直接看課本就能找到推導過程。

那麼記住這個公式後，題目中的積分曲線是什麼形式，我們就直接代入這個公式就行了，然後將積分變數的上下限寫上去就可以了。代入的時候需要注意d表示的是求微分的運算就可以了，比如d(2x)=2dx。

舉一個例子吧。

例計算 $[intlimits_L {left( {x + y} ight)ds} ]$ 其中L為

（1）曲線y=x+1上在（0,1）到（1,2）之間的一段弧

（2）曲線x=2y-1上在（-1,0）到（1,2）之間的一段弧

解（1）

（2）

大家看，根本不用死記硬背那三種情況的計算公式，直接代入是不是很方便呢？！

二、第二類曲線積分的計算記憶方法

第二類曲線積分計算的更簡單，我告訴大家，根本不用去記憶上面的公式。怎麼計算呢？四個字：直接代入。

也就是說，只需要將曲線方程直接代入積分表達式，是誰，就把積分積分表達式里的這個變數全部替換即可。但是要注意最後是起點為積分上限，終點為積分下限。下面舉例說明。

例計算曲線積分 $[intlimits_L {{y^2}dx + 2xydy} ]$ 其中L為

（1）拋物線 $[y = {x^2}]$ 上從（0,0）到（1,1）的一段弧

（2）拋物線 $[x = {y^2}]$ 上從（1,1）到（0,0）的一段弧

解：（1）

因為曲線以 $[yleft( x ight)]$ 的形式給出，所以直接代入曲線形式，將積分裡面所有的y都替換成 $[{x^2}]$

，包括dy中的y（但要注意這個時候要求微分）。最後再確定好積分上下限積分。如下：

$[egin{array}{l} intlimits_L {{y^2}dx + 2xydy} = intlimits_L {{{left( {{x^2}} ight)}^2}dx + 2xleft( {{x^2}} ight)dleft( {{x^2}} ight)} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} \ {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} = intlimits_L {5{x^4}dx} \ {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} = int_0^1 {5{x^4}dx} = 1 end{array}]$

（2）

因為曲線以 $[xleft( y ight)]$ 的形式給出，所以直接代入曲線形式，將積分裡面所有的x都替換成 $[{y^2}]$ ，包括dx中的x（但要注意這個時候要求微分）。最後再確定好積分上下限即可。如下：

$[egin{array}{l} intlimits_L {{y^2}dx + 2xydy} = intlimits_L {{y^2}dleft( {{y^2}} ight) + 2{y^2}ydy} \ {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} = intlimits_L {4{y^3}dy} \ {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} = int_1^0 {4{y^3}dy} = - 1 end{array}]$

===============

好了，這篇文章看似篇幅這麼長，其實就說了兩個基本的問題。總結下：

第一類曲線積分計算方法：

（1）記住公式

$[intlimits_L {fleft( {x,y} ight)ds} = intlimits_L {fleft( {x,y} ight)} sqrt {{{left( {dx} ight)}^2} + {{left( {dy} ight)}^2}} ]$

（2）直接代入曲線方程。該求微分求微分，該確定積分上下限就確定積分上下限即可。

第二類曲線積分計算方法：

（1）直接代入曲線方程

（2）確定積分上下限直接計算即可。