
Korean: 
이제 우리 단위 법선 벡터를 생성 하는 방법에 대해 조금 알고 있는
곡선-어떤 시점에서 그건 우리가 마지막 비디오-무슨 짓을 했는지
탐구 하 고 재미 있는 식 시작 싶어.
그래서 식: 닫힌된 루프 주위 정수 선 우리가 긍정적인 카운터에 갈 거 야
시계 방향
벡터 필드의 F를 곡선 ds에 언제 든 지 단위 법선 벡터와 점
(내가 자홍색으로 ds를 쓸 거 야)
ds입니다.
그래서 먼저이 말은 개념 수 있습니다.
그리고 하자 약간 조작 보고 만약 우리가 함께 올 수 있습니다
그리고 흥미로운 결론 우리가 실제로 그것을 조작할 것 이다
그린의 정리를 사용 합니다.
우리는 실제로 거 올라와 2 차원
버전의 발산 정리
복잡 한 모든 소리 아주
하지만 잘하면 우리가 직감의 a little bit을 얻을 수 있습니다.
왜 그것을 위해 그것은 사실 조금의 상식.
그래서, 먼저 그냥 이것에 대해 생각할 수 있습니다.
그래서 하자 여기 좌표 평면을 그릴 날 무승부 하자
그래서 우리는 화이트에서
그래서이 여기에 그 게 통해 우리 y 축 오른쪽
저기 그건 우리의 x 축
스스로 나의 커브 볼 수 있도록 나의 커브를 그려 보자
같은 것-내가 푸른 색에 그것을 할 거 야
그래서 나의 커브는 이런식으로 생겼을 수 있습니다-내 컨투어
긍정적인, 시계 반대 방향으로 고
그냥 그렇게
그리고 지금 우리는 우리의 벡터 필드와 만난다
우리는 여러 번 봤어요
내가 할 수 있는-내 벡터 필드 사용 하 여 벡터를 연결 합니다.
x-y 평면에서
그리고 그것을 것입니다-일부 기능으로 정의할 수 있습니다.
x와 y는 P를
x의 일부 기능 및 y 시간 i 단위 벡터
그래서 그것은 어떤 내가의 구성 요소 벡터 필드는 어떤 x 및 y 지점
다음 어떤 j 구성, 또는 우리가 무엇을 곱하면합니다
j 구성 요소에 의해
또는 x 및 y에 의해 수직 구성 요소
그래서 일부 기능을 x 및 y 나 번
게다가 x의 다른 스칼라 함수 및 y
j 회
그래서 당신은 내게 라도 하는 경우
그걸로 연결된 벡터 될 것입니다.
어떤 시점에 따라 관련된 벡터
어떻게 우리는이 함수를 정의 합니다.
여기에 바로이 식
우리는 선 통합 복용
우리는이 곡선에 대 한 포인트에 대해 구체적으로 관심
이 윤곽선을 따라
여기 고 그렇게 하면 바로 위에 사실은이 대 한 생각
여기에 바로이 작품은 실제로 우리에 게
우리가 전에 이러한 infinitesimally 작은 조각을 모두 정리해
그래서 우리가 걸릴 f 점 n
따라서가이 곡선에 지점에 대 한 그냥 생각
그래서이 곡선의 한 점을 바로 여기에이 수 있습니다.
그래서 거기는 벡터는 연관
그건 벡터 필드 않습니다.
그래서 f 보여주는 그런 바로 저기
그래서 그는 그 시점에서 f 될 수도 있습니다.
그리고 우리는 거 점이 그것 단위 법선 벡터와 그 시점에서
그래서 단위 법선 벡터 그런 뭔가 볼 수 있습니다
그 시점에서 n 모자 것입니다.
그 시점에서 벡터 필드입니다.

Spanish: 
Ahora que sabemos un poco sobre como construir un vector normal unitario
en cualquier punto en una curva - eso lo hicimos en el ultimo video -
Quiero empezar a explorar una expresion interesante.
La expresion es: el integral lineal alrededor de un
orientación hacia la derecha
de un campo vectorial f salpicado el vector normal de la unidad en cualquier punto de dicha curva ds
(Voy a escribir ds en magenta)
DS.
Lo primero permite conceptualizar lo que esto incluso está diciendo
y luego vamos a manipular un poco para ver si podemos llegar con
y conclusión interesante, realmente se manipulan
vamos a usar el teorema de green
y realmente nos vamos a topar con un 2 dimensiones
versión del teorema de la divergencia
que todos los sonidos muy complicados
pero ojala que podamos llegar a un poco de intuición
pues qué es realmente un poco de sentido común.
Así, en primer lugar permite sólo pensar en esto.
Así que permítanme sorteo-let me dibuje un plano de coordenadas aquí
así lo hacemos en blanco
Así que este derecho aquí que es nuestro eje de y
allí está nuestro eje x
Permítanme llamar nosotros mismos mi curva por lo que podría ser mi curva
algo así como--voy a hacer en un color azul
así que mi curva podría ser algo como esto-mi contorno
y su va en la dirección positiva, a la izquierda
sólo así
y ahora tenemos nuestro campo de vector y sólo un recordatorio
hemos visto esto varias veces
Puedo-mi campo vectorial asociará un vector con cualquiera
punto en el plano x-y
y lo hará--puede ser definido como una función
de x e y, que yo llamo que p
alguna función de x y y veces la i vector unitario
por lo que dice lo que el me componente del campo vectorial es para cualquier x e y punto
y, a continuación, a lo que el componente j, o lo multiplicamos
el componente de j por
o punto de la componente vertical de para cualquier x e y
así que alguna función de x e y siempre me
Además de algunas otras funciones escalares de x e y
veces j
y si me dan cualquier punto
habrá un vector asociado con él
cualquier punto de que hay un vector asociado en función
¿Cómo definimos esta función.
Pero esta expresión por aquí
Estamos tomando una línea integral
nos preocupa específicamente sobre los puntos a lo largo de esta curva
a lo largo de este contorno
por aquí y así permite pensar sobre lo que esto es realmente
Esta pieza sobre aquí realmente nos está diciendo
antes nos resumir todas estas piezas infinitesimal
así que si nosotros sólo tomar f punto n
permite tan sólo pensar en un punto de esta curva
así un punto de esta curva puede ser este punto derecho por aquí
asociado hasta que punto allí es un vector
es lo que hace el campo vectorial
Tan f podría ser algo como ese derecho por allí
así podría ser en ese momento f
y vamos a punto con el vector normal de la unidad de ese momento
por lo que el vector normal de la unidad podría ser algo así
en ese momento sería n sombrero
es en ese momento el campo vectorial

Thai: 
ตอนนี้เรารู้นิดหน่อยแล้วว่าจะสร้างเวกเตอร์หน่วยนตั้งฉาก
ณ จุดใด ๆ บนเส้นโค้งยังไง -- นั่นสิ่งที่เราทำในวิดีโอที่แล้ว --
ผมอยากเริ่มศึกษาพจน์ที่น่าสนใจ
พจน์นั้นคือ อินทิกรัลเส้นรอบวงปิด และเราจะใส่ไนทิศบวก
ทวนเข็มนาฬิกา
ของสนามเวกเตอร์ F ดอทกับเวกเตอร์หน่วยตั้งฉาก ณ จุดใด ๆ บนเส้นโค้ง ds
(ผมจะเขียน ds ด้วยสีบานเย็นนะ)
ds
งั้นอย่างแรก ลองดูว่านี่หมายความว่าอะไร
แล้วลองจัดการมันหน่อย เพื่อดูว่าเราจะ
สรุปอะไรดี ๆ ได้ไหม, ที่จริงเราจะลอง
จัดการมันด้วยทฤษฎีของกรีน
แล้วเราจะได้ทฤษฏีไดเวอร์เจนซ์
แบบ 2 มิติ
ซึ่งฟังดูซับซ้อนมาก
แต่หวังว่าเราจะได้สัญชาตญาณนิดหน่อย
ว่าทำไมมันถึงเข้าใจได้
อย่างแรก, ลองคิดถึงนี่กัน
ขอผมวาด -- ขอผมวาดระนาบพิกัดตรงนี้
เราทำด้วยสีขาวนะ
งั้นนี่ตรงนี้คือแกน y
นั่นตรงนี้คือแกน x
ขอผมวาดเส้นโค้งหน่อย เส้นโค้งอาจ
เป็นแบบ -- ผมจะใช้สีฟ้านะ
เส้นโค้งผมอาจมีหน้าตาแบบนี้ -- เส้นระดับ
และมันเป็นบวก, ทิศทวนเข็มนาฬิกา
แบบนั้น
และตอนนี้ เรามีสนามเวกเตอร์, และเพื่อเตือนความจำ
เราเห็นเส้นหลาย ๆ เส้นนี่
ผมสามารถ -- สนามเวกเตอร์ผมจับคู่เวกเตอร์กับ
จุดใด ๆ บนระนาบ x-y
และมันจะ -- มันสามารถนิยามด้วยฟังก์ชัน
ของ x กับ y, ซึ่งผมจะเรียกมันว่า P
ฟังก์ชันของ x กับ y คูณเวกเตอร์หน่วย i
มันก็บอกว่าองค์ประกอบ i ของสนามเวกเตอร์คืออะไร สำหรับจุด x กับ y
แล้วองค์ประกอบ j, หรือสิ่งที่เราคูณ
องค์ประกอบ j
หรือองค์ประกอบแนวดิ่ง สำหรับ x กับ y ใด ๆ
ดังนั้นฟังก์ชันของ x กับ y คูณ i
บวกฟังก์ชันสเกลาร์อีกอัน x กับ y
คูณ j
และดังนั้น หากคุณให้จุดผมมา
มันจะมีเวกเตอร์นึงคู่กัน
จุดใด ๆ จะมีเวกเตอร์อยู่ ขึ้นกับ
ว่าเรานิยามฟังก์ชันนี้ยังไง
แต่พจน์นี้ตรงนี้
เรากำลังหาอินทิกรัลเส้น
เราสนใจเฉพาะจุดที่อยู่บเนเส้นโค้งนี้
ตลอดเส้นนี่
ตรงนี้ และลองคิดดูว่านี่ตรงนี้
ส่วนนี่ตรงนี้กำลังบอกอะไรเรา
ก่อนเรารวมส่วนเล็กจิ๋วทั้งหมดนี่
เราก็ต้องหา f ดอท n
ลองคิดดูถึงจุดนึงบนเส้นโค้งนี่
งั้นจุดนึงบนเส้นโค้ง อาจเป็นจุดนี่ตรงนี้
มันก็มีเวกเตอร์อยู่
นั่นคือสิ่งที่เวกเตอร์สนามบอกมา
ดังนั้น f อาจมีหน้าตาแบบนั่นตรงนั้น
นั่นก็อาจเป็น f ณ จุดนั้น
และเราจะดอทมันกับเวกเตอร์หน่วยตั้งฉาก ณ จุดนั้น
แล้วเวกเตอร์หน่วยตั้งฉากอาจเป็นแบบนั้น
นั้นจะเป็น n ใส่หมวก ณ จุดนั้น
นี่คือสนามเวกเตอร์ ณ จุดนั้น

Portuguese: 
Agora que sabemos um pouco como construir um vetor unitário normal
à qualquer ponto em uma curva.. Isso é o que nós fizemos no último vídeo
Eu quero começar a explorar algumas expressões interessantes.
Então, a expressão é: "Uma pedaço de linha com um círculo fechado", vamos entrar no contador positivo no
Sentido horário
de um campo do vetor F dotado com o vetor unitário normal a qualquer ponto sobre a curva ds
(Vou escrever ds na cor magenta)
ds.
Então, primeiramente, vamos conceitualizar o que isso está dizendo
e então vamos manipulá-lo um pouquinho para ver se podemos apresentá-lo
com uma conclusão interessante. Nós, de fato, vamos manipulá-lo
vamos usar o teorema verde
e nós vamos apresentá-lo em uma versão
bidimensional do teorema divergente
Tudo parece muito complicado
mas ainda bem que podemos usar um pouco de intuição
para isso, à medida que o porquê de tudo é, na verdade, um pouco de senso comum.
Então, primeiro, vamos apenas pensar sobre isso.
Deixe-me desenhar um plano cartesiano aqui
Vamos fazê-lo em branco
Então esse daqui é o nosso Eixo Y
e logo aqui está o nosso Eixo X
Deixe-me desenhar a curva, que deveria se parecer
algo assim. Vou desenhá-la em azul
Então, minha curva deveria parecer algo assim... Meu contorno...
e ela está indo positivamente, no sentido anti-horário
Assim
e agora que temos o campo do vetor, apenas um lembrete:
Nós vimos isso inúmeras vezes
O campo do vetor vai se associar a um vetor em qualquer
ponto sobre o plano X-Y
e ele vai... Ele pode ser definido como uma função
de X e Y, a qual eu vou chamar de P
Uma função de X e Y dá o momento do vetor unitário "i"
então ele diz o que o componente "i" do campo do vetor é para qualquer ponto em X e Y
e então o que o componente "j", ou pelo que multiplicarmos
o componente "j"
Ou a componente vertical por qualquer ponto em X e Y
Algumas funções de X e Y dão o momento de "i"
Mais algumas outras funções escalares de X e Y
dão o momento de "j"
E então se eu tiver qualquer ponto
haverá um vetor associado à ele
Qualquer ponto alí é um vetor associado dependendo
como nós definimos essa função.
Mas nessa expressão aqui
Nós estamos pegando um pedaço de linha...
Nós estamos cuidando especificamente dos pontos durante essa curva
ao longo do contorno bem aqui.
E então vamos pensar sobre como isso está de verdade
Esse pedaço aqui está nos contando.
Antes de nós resumirmos todos esses infinitos pequenos pedaços...
Se nós pegarmos F vezes N
Vamos pensar em um ponto na curva
Então o ponto nessa curva deveria ser aqui
Associando com aquele ponto existe um vetor
que é o que o campo do vetor faz
Então F deveria parecer algo como isso daqui
Isso deveria ser F naquele ponto
e vamos multiplicá-lo com um vetor unitário normal naquele ponto
Então o vetor unitário normal deveria parecer algo assim
que seria uma cobertura N nesse ponto
Esse é o campo do vetor naquele ponto

Serbian: 
Sada kada znamo malo o tome kako napraviti vektor
na bilo kojoj tacki na grafiku -- to smo radili u prosloj lekciji--
Zelim poceti sa istrazivanjem zanimljivih izraza
Izraz glasi : Integral je oko zatvorene petlje i mi cemo ici u pozitivnom skalaru.
Orijentacija okretanja kazaljke na satu
vektora oznacenim sa "F" sa normalnim vektorom na bilo kojoj tacki petlje ds
(Napisacu ds u ljubicastoj boji)
ds.
Prvo cemo da damo koncept onoga o cemu se prica
i onda cemo to obradjivati da vidimo da li mozemo da se nosimo s time
a interesantno je to da cemo ipak to obradjivati
Koristicemo greenovu teoremu
i zapravo cemo radit isa 2 dimenzije
verzija teoreme divergencije
sto zvuci veoma komplikovano
ali na svu srecu cemo mi to uprostiti
za one koji zapravi shvataju bar malo gradiva
Prvo da razmislimo o ovome.
Nacrtajmo pravougaonik
uradicemo to belom bojojm
tacno je ovde gde je nasa y osa
a ovde je nasa x osa
sada mi dozvolite da nam nacrtam liniju koja ce izgledati
ovako nekako -- Nacrtacu je plavom bojom
Tako da ce moja kriva izgldati ovako
i ona ide u pozitivnom smeru kretanja kazaljki na satu
bas ovako
sada kada imamo nas vekttor, i samo da potsetim
videli smo ovo vise puta
Ovom vektoroskom polju moze da bude dodeljen bilo koji
vektor u delu x-y
i bice -- moze biti definisana kao funkcija
od x i y, koju cu ja nazvati sa "P"
neke funkcije od x i y
...
i onda j komponenta, ili ono sa cime smo pomnozili
komponenty j
ili vertikalna komponenta za bilo koju tacku x y
tako da neke funkcije x y
plus neke ostale skalarne funckije x i y
j
tako da ako mi date bilo koju tacku
za nju ce postojati vektor
za svaku tacku je pripisan vektor koji joj pripada
tako mi definisemo ovu funkciju
ali izraz je zapravo
sto mi pricamo o integralu linije
mi brinime posebno za tacke na ovoj krivoj
oko ove konture
tacno ovde tako da najbolje je da razmislimo sta je zapravo
ovaj deo i sta nam on govori
pre nego sto saberemo sve ove delove
tako da ako uzmemo u obzir samo tacku f
razmislimo samo o tacki na ovoj krivoj
tako da tacka na ovoj krivoj moze da bude bas ovde
pripojena sa tom tackom postoji vektor
to je sta polje vektora radi
tako da f moze da izgleda ovako
...
i mi cemo da ga obelezimo sa tackom normalnog vektora
u toj tacki
tako da normalan vektor treba da izgleda tako
to bi trebalo da bude n tacka
ovo je vektor u toj tacki

French: 
Maintenant que nous savons un peu sur la manière de construire un vecteur normal d'unité

Arabic: 
الآن وبما أننا نعرف القليل عن إنشاء متجه الوحدة العادي
في أي نقطة في المنحنى -- وهذا ما قمنا به في الفيديو السابق--
أريد أن أستكشف و استمتع في التعابير.
التعبير أو الكسر هو : تكامل الخط حول حلقة مغلقة ، وسنتجه نحو عكس عقارب
الساعة الموجب
لحقل المتجه F منقط بمتجه الوحدة العادي في أي نقطة على المنحنى ds
(سأكتب ds باللون الأرجواني )
ds.
أولاً لنصوّر ما نفهمه من هذا
و من ثم نتلاعب بها قليلاً لنرى ما بإمكاننا الوصول إليه
من حقائق مدهشة ، في الحقيقة سنتلاعب بها
وسنستخدم نظرية غرين
وسنتوصل إلى نسخة ثنائية الأبعاد
من نظرية التباعد
والتي تبدو معقدة
ولكن نأمل أن يحالفنا الحدس
لأنها تبدو منطقية.
إذاً ، أولاً لنفكر بشأن هذا.
دعوني أرسم - دعوني أرسم لوح إحداثيات هنا
باللون الأبيض
هذا هو محور الصادات
وهذا محور السينات
دعوني أرسم المنحنى الخاص بنا والذي قد يبدو
ربما هكذا -- سأقوم بذلك باللون الأزرق
المنحنى قد يبدو هكذا - عكس
وسيكون في الاتجاه الموجب ، عكس اتجاه عقارب الساعة
هكذا
الآن لدينا المتجه ممتلئ ، ولتذكيركم
فلقد رأينا هذا عدة مرات
أستطيع - المتجه ممتلئ الخاص بي سيرتبط مع أ]
نقطة على لوح افحداثيات
وسيـ -- يمكن ان نعرفه كدالة ما
بدلالة X و y , وسأسمي ذلك P
دالة x و Y ضرب i متجه الوحدة
وذلك يبين ماهية العنصر i للمتجه الممتلئ لأي نقطة x وy
وثم ماهية العنصر j او ما نضرب
به العنصر j
أو أي عنصر عمودي عليه لأي نقطة x وy
إذاً دالة ما بدلالة x و y ضرب i
زائد أي دالة مدرجة بدلالة x و y
ضرب j
وإذا أعطيتني أي نقطة
سيكون هناك متجه مرتبط بها
لأي نقطة يوجد هناك متجه مرتبط يعتمد على
آلية تعريف هذه الدالة
ولكن هذا التعبير هنا
هو أننا نأخذ تكامل خط
نحن نهتم تحديداً بالنقطة على هذا المنحنى
وعلى امتداد هذا المعكوس
هنا ، ولنفكر ماهي حقيقة هذا الشيء
هذه القطعة هنا تخبرنا
قبل أن نجمع كل هذه القطع الصغيرة المتناهية في الصغر
إذا أخذنا f ضرب n نقطياً
لنفطر فقط بشأن نقطة على هذا المنحنى
نقطة على هذا المنحنى ربما تكون هذه النقطة هنا
مرتبطة بالنقطة هنا وهو متجه
هذا ما يقوم به المملوء به المتجه
إذا ً f ربما تكون شيئاً مشابهاً لهاذ هناك
إذاً هذا ربما يكون f عند هذه النقطة
وسنقوم بضربها نقطياً بمتجه وحدة عادي عند هذه النقطة
لذا فإن متجه الوحدة العادي قد يبدو مشابهاً لهذا
هذا سيكون n شرطة عند تلك النقطة
هذا متجه ممتلئ عند هذه النقطة

English: 
- [Instructor] Now that
we know a little bit about
how to construct a unit normal vector
at any point in a curve,
that's what we did in the last video,
I want to start exploring
an interesting expression.
So the expression is the line
integral around a closed loop,
and we're gonna go in the positive,
counterclockwise orientation,
of a vector field F,
dotted with the unit normal vector
at any point on that curve ds,
I'll write ds in magenta, ds.
So first let's conceptualize
what this is even saying,
and then let's manipulate it a little bit
to see if we can come up with
and interesting conclusion.
We actually will manipulate it.
we'll use Green's theorem,
and we're gonna actually come up
with a two-dimensional version
of the divergence theorem,
which all sounds very complicated,
but hopefully we can get a
little bit of an intuition for it
as to why it actually a
little bit of common sense.
So first, lets just think about this.
So let me draw a coordinate plane here.

Urdu: 
اب چونکہ ہم تھوڑا بہت جان چکے ہیں کہ کیسے بنایا جاتا ہے اکائی عام ویکٹر
قوس میں کسی بھی مقام پر -- جو کہ ہم نے پچھلی ویڈیو میں دیکھا--

English: 
So let me do it in white.
So this right over
here, that's our y-axis.
That over there is our x-axis.
Let me draw ourselves my curve.
So my curve might look something like...
I'll do it in the blue color.
So my curve might look
something like this, my contour,
and its going in the positive,
counterclockwise direction just like that.
And now we have our vector field,
and, just a reminder, we've
seen this multiple times.
My vector field will associate a vector
with any point on the x-y plane,
and it can be defined as
some function of x and y,
actually, I'll call that P,
some function of x and y
times the i unit vector.
So it says what the i component
of the vector field is
for any x and y point,
and then what the j component,
or what we multiply the j component by
or the vertical component
by for any x and y point.
So some function of x and y times i

English: 
plus some other scalar
function of x and y times j.
And so, if you give me any point,
there'll be an associated vector with it.
Any point, there's an associated vector,
depending on how we define this function.
But this expression right over here,
we're taking a line integral.
We care specifically about
the points along this curve,
along this contour right over here.
And so let's think about what
this piece right over here
is actually telling us before we sum up
all of these infinitesimally small pieces.
So if we just take F dot n,
so let's just think about
a point on this curve.
A point on this curve, maybe
this point right over here.
So associated with that
point, there is a vector.
That's what the vector field does.
F might look something
like that right over there.
So that might be F at that point,
and then we're gonna dot it
with the unit normal vector at that point.
So the unit normal vector
might look something like that.
That would be n hat at that point.

English: 
This is the vector field at that point.
When you take the dot product,
you get a scaler quantity.
You essentially just get a
number and you might remember it
and there's several
videos where we go into
detail about this but
that tells you how much
those two vectors go together.
It's essentially, if they're
completely orthogonal
to each other you're gonna
get zero and if they go
completely in the same
direction it's essentially
just gonna multiply their
magnitudes times each other.
And since you have a
unit normal vector here,
what this is essentially going
to give you is the magnitude
of the vector field F that
goes in the normal direction.
So, think of it this way.
So let's think about the component of this
that goes in the normal direction.
It might look something
like that, this would be
the component that goes
the tangential direction.
So, this expression right
over here is going to give us
the magnitude of this vector.
Let me write this down,
this right over here is the

English: 
magnitude of the component
of F, in the normal direction
or in the same direction
as that unit normal vector.
Normal direction.
And then were gonna multiply
that times a very infinitely
small length of our contour,
of our curve right around that point.
So were gonna multiply that thing,
times this right over here.
And so you might say, "Well,
you know, okay, I kind of
"get what that is saying but
"how could this ever
be physically relevant,
"or what could be the intuition for what
"this expression is even measuring?"
And to think about
that, I always visualize
this is in two dimensions.
You will later do things like this
in three dimensions, so imagine this is
a two dimensional universe
and we're studying gases.
And so you have all these gas
particles in a two dimensional
universe so they only can have
kind of an x and y coordinate
and this vector field is essentially
telling you the velocity
at any point there.

English: 
So this is the velocity of
the particles at this point,
this is the velocity of
particles at that point,
that is the velocity of
particles at that point.
And so, when you take F right
on this curve, that's the
velocity of the particles at that point
they're going in that direction.
When you dot it with N it tells you,
essentially, the speed going straight out
right at that point, and then when
you multiply that times ds,
you're essentially saying,
at any given moment, how
fast, or at that point
right over there on the curve,
how fast are the particles
exiting the curve?
And so if you were to sum up all of them
which is essentially what this
integral is doing with the
slight integral, it's essentially saying,
how fast are the particles
exiting this contour?
Are they even entering
the contour if you get
a negative number?
But since were taking the
unit normal vector that
is outward-pointing, it's saying,
"How fast are they exiting this thing?"
If you've got a negative
number, that means
there might be some net entrance.

English: 
So, this whole expression
is if you take that analogy,
it doesn't have to have that
physical representation,
how fast are particles, are
two dimensional gas particles
exiting the contour?
In the future, you can do three dimensions
where you have a surface and you can say,
"How fast are things
exiting this surface?"
And so let's start, now
that we hopefully have
a decent conceptual understanding of what
this could represent let's play
around with it a little bit,
especially because we know
how to define a normal vector.
So, let's rewrite it using what we know
about how to construct a normal vector.
So, if we rewrite it our
integral becomes this.
We have our vector field F,
dotted with the normal vector.
The normal vector we can write this way.
A normal vector we saw was
dy times y minus dx times j.
And then we had to divide
it by its magnitude.

English: 
In order to make it a unit normal vector,
the magnitude was this right over here,
dx squared by dy squared,
which is the same thing
as ds which is the same thing
as that little mini arc length
that infinitely small
length of our contour.
So, we're going to divide it by ds.
We're going to divide it by ds.
And then we're going to
multiply it times ds.
We're going to multiply it by ds.
And ds is just a scanner quality,
and so we can actually even multiply
this thing times ds before
taking the dot product
or vice versa.
But these two things
are going to cancel out.
And so we're essentially
left with F dot this thing
right over here.
But we have F defined
right over here so let's
take the dot products.
I'll just write the line
integral symbol again.
We're going in the
counterclockwise direction.
And when we evaluate and now let's in--
let me pick a color that I have not used.
I may have used many colors
so I'll do yellow again.

English: 
So now, let's evaluate
F dot this business.
So, dot product fairly straightforward.
You take the product of the x-components,
or essentially the magnitude
of the x-components.
So, it's going to be P of
Y, P of xy, times the dy,
plus the product of the
magnitudes of the y component
or the j component.
So it's going to be plus
Q of xy times minus dx,
times negative dx.
Well, that's going to give us
negative Q of xy,
negative Q of xy times dx.
So, this is kind of interesting statement.
We've seen something not too
different from this before,
when we saw even just the
definition of Green's theorem.
Let me rewrite here just so we remember.
So, the definition when we
learned about Green's theorem,

English: 
it told us if we're taking a
line integral over this contour
and there's multiple ways
to write it, but one way
that we often still see it,
and we've explored it already
in our videos is if we were
to say M times dx plus N
times dy, this is equal to--
and I'm just restating
Green's theorem right here--
this is equal to the double
integral over the region
that this contour surrounds of...
And whatever function
you have here times dy,
you take the partial of
that with respect to X,
the partial of N with respect to X,
and from that we subtract
whatever was on the dx side,
so the partial of M with
respect to Y, and then
you could say times df,
dy, or you could say da,
for the infinitesimally
small little chunk of area.
So, I'm going to write DA here.

English: 
All right, dx, dy, or it could be dy.
Or we might as well write DA here,
since we're speaking in generalizations,
where DA is an infinitely
small little chunk of area.
So this here, this is just a
restatement of Green's theorem
so we already know this.
This is a restatement of Green's Theorem
and how can we apply it here?
Well, it's the same thing.
You have a little bit
of sine differences but you
can apply Green's theorem here.
This is going to be equal
to the double integral
over the region that
this contour surrounds.
And then, what we want to do is,
we want to look at whatever
is the function that
is being multiplied by the dy.
In this case, this is the function
that is being multiplied times the dy.
And we want to take the partial
of that in respect to X.
So, we're going to take the
partial of P with respect to X.
So, the partial of P with respect to X.

English: 
And then from that we
are going to subtract.
We're going to subtract
the other function,
whatever's being multiplied by dx,
we're going to take the partial
of that with respect to Y.
So here, we're going to take the partial
of this whole thing with respect to Y,
but we have a negative out here
so it's going to be minus partial of Q
with respect to Y and then we have DA,
and obviously, these two negatives,
subtracting the negative
just gives you a positive.
So this is going to be
equal to the double integral
over the region.
And maybe you already
see where this is going,
maybe getting a little bit excited.
The partial of P with respect to X plus
the partial of Q with respect to Y, dA.
Now, okay, I took the partial of...
"Well, what is this telling me, Sal?"
Well, look at this thing right over here.
P was originally the
function that was telling us
the magnitude in the X direction.
Q was telling us the
magnitude of the Y direction.
We're taking the partial
of this with respect to X,
we're taking the partial
of this with respect

English: 
to Y and we're summing them.
This is exactly the divergence of F.
And if that doesn't make any sense,
go watch the video on divergence.
This right over here
is the divergence of F.
This is the divergence
by definition, really,
this is the divergence
of our vector field F.
And so, we have a very interesting thing.
This thing that we saw was
the original expression
that we started studying
which is essentially saying,
"What's the speed at which the particles
"are exiting this surface?"
We now get it in terms of
this little expression.
And we'll interpret in an intuitive way
and a little bit of this in a little bit.
So this is equal to the
divergent of F times dA
over the whole region
so the double integral.
So, we're summing up the
divergence of F times
the infinite, really small
little chunk over there
and we're summing them
up over the whole region.
Now, why does that make intuitive sense?

English: 
And for you to realize why
it makes intuitive sense,
you just have to remind
yourself what the divergence is.
Divergence is a measure of
whether things are expanding
or diverging or kind of contracting.
If you have a point over
here where around that,
the particles are kind of
moving away from each other,
you would have a positive divergence here.
If you have a point,
sometimes called a sync,
where all of the particles
are kind of condensing
or converging you would have
a negative divergence here.
And now, so this should
make a lot of sense.
You take any little
infinitely small area in this,
you take any little infinitely small area,
and then multiply that
times the divergence there.
So, the more divergence, you're going
to get a larger number and then you sum
it up across the entire region.
That makes a lot of sense.
The more diverging that's
going on through your region,
obviously, the more stuff that's going
to be exiting your boundary over here.

English: 
So, it actually makes complete sense,
or hopefully it makes a
little bit of complete sense.
If you were to see how fast are things
exiting the surface, it's
really the two dimensional flux,
how fast are things exiting the surface,
and you take the sum of all of them,
that's going to be the
same thing of summing
all of the divergences over this area
that the contour is surrounding.
So, hopefully that makes a
little bit of sense to you.
And it's actually another way of
kind of thinking about Green's Theorem.
And we also have just
explored what we just said,
this expression that the
divergence summed over
this region over here is the
same thing as the F dot n
over the contour.
That is, essentially, the two dimensional
divergence theorem.
