DEFINIÇÃO
Seja F : R 3 → R 3 F: \R^3 \to \R^3 F : R 3 → R 3 C 1 C^1 C 1
Definimos o rotacional de F F F rot F \rot F rot F ∇ × F \nabla \times F ∇ × F
rot F = ∇ × F = ∣ e 1 e 2 e 3 ∂ ∂ x ∂ ∂ y ∂ ∂ z F 1 F 2 F 3 ∣ = ( ∂ F 3 ∂ y − ∂ F 2 ∂ z , − ∂ F 3 ∂ x + ∂ F 1 ∂ z , ∂ F 2 ∂ x − ∂ F 1 ∂ y ) \begin{aligned}
\rot F = \nabla \times F &= \begin{vmatrix}
e_1 & e_2 & e_3\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
F_1 & F_2 & F_3
\end{vmatrix}\\
&=\left(\frac{\partial F_3}{\partial y}- \frac{\partial F_2}{\partial z}, -\frac{\partial F_3}{\partial x} + \frac{\partial F_1}{\partial z}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right)
\end{aligned} rot F = ∇ × F = e 1 ∂ x ∂ F 1 e 2 ∂ y ∂ F 2 e 3 ∂ z ∂ F 3 = ( ∂ y ∂ F 3 − ∂ z ∂ F 2 , − ∂ x ∂ F 3 + ∂ z ∂ F 1 , ∂ x ∂ F 2 − ∂ y ∂ F 1 ) Como simples exemplos, podemos tomar os campos vetoriais F 1 F_1 F 1 F 2 F_2 F 2
F 1 ( x , y , z ) = ( − y , x , z ) F 2 ( x , y , z ) = ( x 2 , 3 x 2 , y + z ) \begin{array}{ll}
F_1(x,y,z) = (-y,x,z) & F_2(x,y,z)=(x^2, 3x^2, y+z)
\end{array} F 1 ( x , y , z ) = ( − y , x , z ) F 2 ( x , y , z ) = ( x 2 , 3 x 2 , y + z ) e de seguida calcular os seus rotacionais:
rot F 1 = ∣ e 1 e 2 e 3 ∂ ∂ x ∂ ∂ y ∂ ∂ z − y x z ∣ = ( 0 , 0 , 1 + 1 ) = ( 0 , 0 , 2 ) rot F 2 = ∣ e 1 e 2 e 3 ∂ ∂ x ∂ ∂ y ∂ ∂ z x 2 3 x 2 y + z ∣ = ( 1 , 0 , 6 x ) \begin{aligned}
\rot F_1 &= \begin{vmatrix}
e_1 & e_2 & e_3\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
-y & x & z
\end{vmatrix}
= (0,0, 1+1) = (0,0,2)\\
\rot F_2 &= \begin{vmatrix}
e_1 & e_2 & e_3\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
x^2 & 3x^2 & y+z
\end{vmatrix}
=(1,0,6x)
\end{aligned} rot F 1 rot F 2 = e 1 ∂ x ∂ − y e 2 ∂ y ∂ x e 3 ∂ z ∂ z = ( 0 , 0 , 1 + 1 ) = ( 0 , 0 , 2 ) = e 1 ∂ x ∂ x 2 e 2 ∂ y ∂ 3 x 2 e 3 ∂ z ∂ y + z = ( 1 , 0 , 6 x )
Seja um campo vetorial F ∈ C 2 F \in C^2 F ∈ C 2 F : R 3 → R 3 F: \R^3 \to \R^3 F : R 3 → R 3 rot F : R 3 → R 3 \rot F: \R^3 \to \R^3 rot F : R 3 → R 3
div ( rot F ) = 0 \ondiv(\rot F) = 0 div ( rot F ) = 0 Demonstração rot F = ( ∂ F 3 ∂ y − ∂ F 2 ∂ z , − ∂ F 3 ∂ x + ∂ F 1 ∂ z , ∂ F 2 ∂ x − ∂ F 1 ∂ y ) \rot F =\left(\frac{\partial F_3}{\partial y}- \frac{\partial F_2}{\partial z}, -\frac{\partial F_3}{\partial x} + \frac{\partial F_1}{\partial z}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) rot F = ( ∂ y ∂ F 3 − ∂ z ∂ F 2 , − ∂ x ∂ F 3 + ∂ z ∂ F 1 , ∂ x ∂ F 2 − ∂ y ∂ F 1 ) div ( rot F ) = ∂ ∂ x ( ∂ F 3 ∂ y − ∂ F 2 ∂ z ) + ∂ ∂ y ( − ∂ F 3 ∂ x + ∂ F 1 ∂ z ) + ∂ ∂ z ( ∂ F 2 ∂ x − ∂ F 1 ∂ y ) = 0 \begin{aligned}
\ondiv(\rot F) &= \frac{\partial}{\partial x} \left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}\right)\\
&+ \frac{\partial}{\partial y} \left(-\frac{\partial F_3}{\partial x} + \frac{\partial F_1}{\partial z}\right)\\
&+ \frac{\partial}{\partial z} \left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right)\\
&= 0
\end{aligned} div ( rot F ) = ∂ x ∂ ( ∂ y ∂ F 3 − ∂ z ∂ F 2 ) + ∂ y ∂ ( − ∂ x ∂ F 3 + ∂ z ∂ F 1 ) + ∂ z ∂ ( ∂ x ∂ F 2 − ∂ y ∂ F 1 ) = 0 Se div G ≠ 0 \ondiv G \ne 0 div G = 0 G G G
Se rot F = ( 0 , 0 , 0 ) \rot F = (0,0,0) rot F = ( 0 , 0 , 0 ) F F F fechado , pois
∂ F 3 ∂ y = ∂ F 2 ∂ z ∂ F 3 ∂ x = ∂ F 1 ∂ z ∂ F 2 ∂ x = ∂ F 1 ∂ y \begin{array}{lll}
\frac{\partial F_3}{\partial y} = \frac{\partial F_2}{\partial z} &
\frac{\partial F_3}{\partial x} = \frac{\partial F_1}{\partial z} &
\frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y}
\end{array} ∂ y ∂ F 3 = ∂ z ∂ F 2 ∂ x ∂ F 3 = ∂ z ∂ F 1 ∂ x ∂ F 2 = ∂ y ∂ F 1
Se rot F = 0 \rot F = 0 rot F = 0 F F F F ⃗ \vec F F gradiente (e conservativo ).
Considerando uma superfície orientada S S S R 3 \R^3 R 3 ∂ S \partial S ∂ S
Podemos definir uma orientação para o bordo através da Regra da Mão Direita
Colocando a mão direita com a palma voltada para a superfície e com o polegar a apontar no sentido da normal,
ficamos com os dedos a apontar na orientação do bordo da superfície.
Podemos tomar como exemplos as superfícies S S S T T T
S = { x 2 + y 2 + z 2 = 1 , z > 0 } , n ⃗ z > 0 S = \{x^2 + y^2 + z^2 = 1, z > 0 \}, \quad \vec n_z > 0 S = { x 2 + y 2 + z 2 = 1 , z > 0 } , n z > 0
T = { x 2 + y 2 = 1 , 0 ≤ z ≤ 1 } , normal exterior T = \{x^2 + y^2 = 1 , 0 \leq z \leq 1 \}, \quad \text{normal exterior} T = { x 2 + y 2 = 1 , 0 ≤ z ≤ 1 } , normal exterior
DEFINIÇÃO
Seja S ⊂ R 3 S \subset \R^3 S ⊂ R 3 F F F C 1 C^1 C 1
∫ S rot F ⋅ n ⃗ = ∫ ∂ S F ⃗ d g \int_S \rot F \cdot \vec n = \int_{\partial S} \vec F \d g ∫ S rot F ⋅ n = ∫ ∂ S F d g em que ∂ S \partial S ∂ S regra da mão direita .
Exemplos Considere a superfície S S S n ⃗ \vec n n S S S
S = { x = − 1 + y 2 + z 2 , x ≤ 0 } F = ( x y , z e x , − y ) \begin{array}{ll}
S = \{ x = -1 + y^2 + z^2, x \leq 0 \} & F = (xy, ze^x, -y)
\end{array} S = { x = − 1 + y 2 + z 2 , x ≤ 0 } F = ( x y , z e x , − y ) Qual o valor de ∫ S rot F ⋅ n ⃗ \int_S \rot F \cdot \vec n ∫ S rot F ⋅ n
A normal está apontada para fora da taça.
∂ S = { x = 0 , y 2 + z 2 = 1 } \partial S = \{ x = 0, y^2 + z^2 = 1 \} ∂ S = { x = 0 , y 2 + z 2 = 1 }
∫ S rot F ⋅ n ⃗ = ∫ ∂ S F ⃗ d g = ∫ ∂ S ( 0 , z , − y ) d g \int_S \rot F \cdot \vec n = \int_{\partial S} \vec F \d g = \int_{\partial S} (0, z , -y) \d g ∫ S rot F ⋅ n = ∫ ∂ S F d g = ∫ ∂ S ( 0 , z , − y ) d g Parametrizar:
{ x = 0 y = cos t z = sin t \begin{cases}
x = 0\\
y = \cos t\\
z = \sin t
\end{cases} ⎩ ⎨ ⎧ x = 0 y = cos t z = sin t g ( t ) = ( 0 , cos t , sin t ) g ′ ( t ) = ( 0 , − sin t , cos t ) F ⃗ ( g ( t ) ) = ( 0 , sin t , − cos t ) F ⃗ ( g ( t ) ) ⋅ g ′ ( t ) = − sin 2 t − cos 2 t = − 1 \begin{array}{ll}
g(t) = (0, \cos t, \sin t) & g'(t) = (0, -\sin t, \cos t)\\
\vec F (g(t)) = (0, \sin t, -\cos t) & \vec F(g(t)) \cdot g'(t) = -\sin^2 t - \cos^2 t = -1
\end{array} g ( t ) = ( 0 , cos t , sin t ) F ( g ( t )) = ( 0 , sin t , − cos t ) g ′ ( t ) = ( 0 , − sin t , cos t ) F ( g ( t )) ⋅ g ′ ( t ) = − sin 2 t − cos 2 t = − 1 Como a orientação da parametrização é a errada,
∫ ∂ S F ⃗ d g = − ∫ 0 2 π − 1 d t = 2 π \int_{\partial S} \vec F \d g = - \int_0^{2\pi} -1 \d t = 2\pi ∫ ∂ S F d g = − ∫ 0 2 π − 1 d t = 2 π Podemos fazer a seguinte observação:
Se F = rot A F = \rot A F = rot A S S S Teorema de Stokes , temos que
∫ S F ⋅ n ⃗ = ∫ S rot A ⋅ n ⃗ = ∫ ∂ S A ⋅ d g ⃗ = 0 \int_S F \cdot \vec n = \int_S \rot A \cdot \vec n = \int_{\partial S} A \cdot \d \vec g = 0 ∫ S F ⋅ n = ∫ S rot A ⋅ n = ∫ ∂ S A ⋅ d g = 0 DEFINIÇÃO
Seja A ⊂ R n A \subset \R^n A ⊂ R n A A A x 0 ∈ A x_0 \in A x 0 ∈ A x 0 x_0 x 0 A A A
Qualquer conjunto aberto estrelado é um conjunto aberto simplesmente conexo (mas o contrário não é sempre verdade).
Exemplos:
TEOREMA
Se G : D ⊂ R 3 → R 3 G: D \subset \R^3 \to \R^3 G : D ⊂ R 3 → R 3 div G = 0 \ondiv G = 0 div G = 0 G G G F : R 3 → R 3 F: \R^3 \to \R^3 F : R 3 → R 3
Tomando G : R 3 → R 3 G: \R^3 \to \R^3 G : R 3 → R 3 div G = 0 \ondiv G = 0 div G = 0 R 3 \R^3 R 3 G = rot F G = \rot F G = rot F
Como podemos calcular o potencial vetor de F F F
Fazer um sistema com as componentes da definição de rotacional, igualadas ao valor conhecido do rotacional, isto é, G G G
r o t F = ∇ × F = ∣ e 1 e 2 e 3 ∂ ∂ x ∂ ∂ y ∂ ∂ z F 1 F 2 F 3 ∣ = ( ∂ F 3 ∂ y − ∂ F 2 ∂ z , − ∂ F 3 ∂ x + ∂ F 1 ∂ z , ∂ F 2 ∂ x − ∂ F 1 ∂ y ) rot F = \nabla \times F = \begin{vmatrix}
e_1 & e_2 & e_3\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
F_1 & F_2 & F_3
\end{vmatrix}
= \left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, - \frac{\partial F_3}{\partial x} + \frac{\partial F_1}{\partial z}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) ro tF = ∇ × F = e 1 ∂ x ∂ F 1 e 2 ∂ y ∂ F 2 e 3 ∂ z ∂ F 3 = ( ∂ y ∂ F 3 − ∂ z ∂ F 2 , − ∂ x ∂ F 3 + ∂ z ∂ F 1 , ∂ x ∂ F 2 − ∂ y ∂ F 1 ) { ∂ F 3 ∂ y − ∂ F 2 ∂ z = G 1 ∂ F 1 ∂ z − ∂ F 3 ∂ x = G 2 ∂ F 2 ∂ x − ∂ F 1 ∂ y = G 3 \begin{cases}
\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} = G_1\\
\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} = G_2\\
\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = G_3
\end{cases} ⎩ ⎨ ⎧ ∂ y ∂ F 3 − ∂ z ∂ F 2 = G 1 ∂ z ∂ F 1 − ∂ x ∂ F 3 = G 2 ∂ x ∂ F 2 − ∂ y ∂ F 1 = G 3
Se G = rot F G = \rot F G = rot F ϕ : R 3 → R \phi: \R^3 \to \R ϕ : R 3 → R
rot ( F + ∇ ϕ ) = rot F + rot ( ∇ ϕ ) = rot F + 0 = G \rot (F + \nabla \phi) = \rot F + \rot (\nabla \phi) = \rot F + 0 = G rot ( F + ∇ ϕ ) = rot F + rot ( ∇ ϕ ) = rot F + 0 = G
Seja ϕ = − ∫ F 1 ( x , y , z ) d x \phi = - \int F_1 (x,y,z) \d x ϕ = − ∫ F 1 ( x , y , z ) d x ϕ \phi ϕ F 1 F_1 F 1 x x x
rot ( F + ∇ ϕ ) = G \rot(F + \nabla \phi) = G rot ( F + ∇ ϕ ) = G A primeira coordenada de F + ∇ ϕ F + \nabla \phi F + ∇ ϕ F 1 + ∂ ϕ ∂ x = F 1 − F 1 = 0 F_1 + \frac{\partial \phi}{\partial x} = F_1 - F_1 = 0 F 1 + ∂ x ∂ ϕ = F 1 − F 1 = 0
Logo, existe um potencial vetor com a primeira componente nula.
É mais fácil perceber através de um exemplo:
Exemplo Seja o campo vetorial F ( x , y , z ) = ( x e y , − 2 e y , z e y ) F(x,y,z) = (x e^y, -2e^y, ze^y) F ( x , y , z ) = ( x e y , − 2 e y , z e y )
Será que F F F
div F = e y − 2 e y + e y = 0 \ondiv F = e^y - 2e^y + e^y = 0 div F = e y − 2 e y + e y = 0 Domínio de F = R 3 F = \R^3 F = R 3
Logo, F F F
Qual o valor de A A A F = rot A F = \rot A F = rot A
Tomando, por exemplo, A 2 A_2 A 2
{ ∂ A 3 ∂ y − ∂ A 2 ∂ z = x e y ∂ A 1 ∂ z − ∂ A 3 ∂ x = − 2 e y ∂ A 2 ∂ x − ∂ A 1 ∂ y = z e y ⇔ { ∂ A 3 ∂ y = x e y − ∂ A 1 ∂ y = − z e y ⇔ { A 3 = ∫ x e y d y = x e y + C 1 ( x , z ) ( − e y + ∂ C 2 ∂ z ) − ( e y + ∂ C 1 ∂ x ) = − 2 e y A 1 = − z e y + C 2 ( x , z ) ⇔ { − − 2 e y + ∂ C 2 ∂ z − ∂ C 1 ∂ x = − 2 e y − ⇔ { − ∂ C 2 ∂ z = ∂ C 1 ∂ x − \begin{aligned}
&\begin{cases}
\frac{\partial A_3}{\partial y} - \frac{\partial A_2}{\partial z} = x e^y\\
\frac{\partial A_1}{\partial z} - \frac{\partial A_3}{\partial x} = -2e^y\\
\frac{\partial A_2}{\partial x} - \frac{\partial A_1}{\partial y} = z e^y
\end{cases}\\
\Leftrightarrow &
\begin{cases}
\frac{\partial A_3}{\partial y} = x e^y\\
\huge -\\
\frac{\partial A_1}{\partial y} = - z e^y
\end{cases}\\
\Leftrightarrow &
\begin{cases}
A_3 = \int x e^y \d y = x e^y + C_1(x,z)\\
(-e^y + \frac{\partial C_2}{\partial z}) - (e^y + \frac{\partial C_1}{\partial x}) = -2 e^y\\
A_1 = -z e^y + C_2(x,z)
\end{cases}\\
\Leftrightarrow &
\begin{cases}
\huge -\\
-2e^y + \frac{\partial C_2}{\partial z} - \frac{\partial C_1}{\partial x}= -2 e^y\\
\huge -
\end{cases}\\
\Leftrightarrow &
\begin{cases}
-\\
\frac{\partial C_2}{\partial z} = \frac{\partial C_1}{\partial x}\\
\huge -
\end{cases}
\end{aligned} ⇔ ⇔ ⇔ ⇔ ⎩ ⎨ ⎧ ∂ y ∂ A 3 − ∂ z ∂ A 2 = x e y ∂ z ∂ A 1 − ∂ x ∂ A 3 = − 2 e y ∂ x ∂ A 2 − ∂ y ∂ A 1 = z e y ⎩ ⎨ ⎧ ∂ y ∂ A 3 = x e y − ∂ y ∂ A 1 = − z e y ⎩ ⎨ ⎧ A 3 = ∫ x e y d y = x e y + C 1 ( x , z ) ( − e y + ∂ z ∂ C 2 ) − ( e y + ∂ x ∂ C 1 ) = − 2 e y A 1 = − z e y + C 2 ( x , z ) ⎩ ⎨ ⎧ − − 2 e y + ∂ z ∂ C 2 − ∂ x ∂ C 1 = − 2 e y − ⎩ ⎨ ⎧ − ∂ z ∂ C 2 = ∂ x ∂ C 1 − Escolhendo C 1 = C 2 = 0 C_1 = C_2 = 0 C 1 = C 2 = 0 A = ( − z e y , 0 , x e y ) A = (-z e^y, 0, xe^y) A = ( − z e y , 0 , x e y )
Considerando S = { x 2 + z 2 = y 2 : 1 < y < 2 } S = \{x^2 + z^2 = y^2: 1 < y < 2\} S = { x 2 + z 2 = y 2 : 1 < y < 2 } n ⃗ y < 0 \vec n_y < 0 n y < 0 ∫ S F ⋅ n ⃗ \int_S F \cdot \vec n ∫ S F ⋅ n
Pelo Teorema de Stokes , sabemos que
∫ S F ⋅ n ⃗ = ∫ S rot A ⋅ n ⃗ = ∫ ∂ S A d g = ∫ C 1 A d g + ∫ C 2 A d g \int_S F \cdot \vec n = \int_S \rot A \cdot \vec n = \int_{\partial S} A \d g = \int_{C_1} A \d g + \int_{C_2} A \d g ∫ S F ⋅ n = ∫ S rot A ⋅ n = ∫ ∂ S A d g = ∫ C 1 A d g + ∫ C 2 A d g C 1 : x 2 + z 2 = 1 , y = 1 C_1: x^2+z^2 = 1, y = 1 C 1 : x 2 + z 2 = 1 , y = 1
g ( t ) = ( cos t , 1 , sin t ) g ′ ( t ) = ( − sin t , 0 , cos t ) A ( g ( t ) ) = ( − e sin t , 0 , e cos t ) \begin{array}{ll}
g(t) = (\cos t, 1, \sin t) & g'(t) = (-\sin t, 0, \cos t)\\
A(g(t)) = (-e \sin t, 0, e\cos t) &
\end{array} g ( t ) = ( cos t , 1 , sin t ) A ( g ( t )) = ( − e sin t , 0 , e cos t ) g ′ ( t ) = ( − sin t , 0 , cos t ) ∫ C 1 A d g = − ∫ 0 2 π A ( g ( t ) ) ⋅ g ′ ( t ) d t = − 2 π e \int_{C_1} A \d g = - \int_0^{2 \pi} A(g(t)) \cdot g'(t) \d t = - 2\pi e ∫ C 1 A d g = − ∫ 0 2 π A ( g ( t )) ⋅ g ′ ( t ) d t = − 2 π e C 2 : x 2 + z 2 = 4 , y = 2 C_2: x^2+z^2 = 4, y = 2 C 2 : x 2 + z 2 = 4 , y = 2
g ( t ) = ( 2 cos t , 2 , 2 sin t ) g ′ ( t ) = ( − 2 sin t , 0 , 2 cos t ) A ( g ( t ) ) = ( − 2 e 2 sin t , 0 , 2 e 2 cos t ) \begin{array}{ll}
g(t) = (2\cos t, 2, 2\sin t) & g'(t) = (-2\sin t, 0, 2\cos t)\\
A(g(t)) = (-2e^2 \sin t, 0, 2e^2\cos t) &
\end{array} g ( t ) = ( 2 cos t , 2 , 2 sin t ) A ( g ( t )) = ( − 2 e 2 sin t , 0 , 2 e 2 cos t ) g ′ ( t ) = ( − 2 sin t , 0 , 2 cos t ) ∫ C 2 A d g = ∫ 0 2 π A ( g ( t ) ) ⋅ g ′ ( t ) d t = ∫ 0 2 π 4 e 2 d t = 8 π e 2 \int_{C_2} A \d g = \int_0^{2 \pi} A(g(t)) \cdot g'(t) \d t = \int_0^{2 \pi} 4 e^2 \d t = 8\pi e^2 ∫ C 2 A d g = ∫ 0 2 π A ( g ( t )) ⋅ g ′ ( t ) d t = ∫ 0 2 π 4 e 2 d t = 8 π e 2 Logo,
∫ S F ⋅ n ⃗ = 8 π e 2 − 2 π e \int_S F \cdot \vec n = 8\pi e^2 - 2\pi e ∫ S F ⋅ n = 8 π e 2 − 2 π e Sejam o campo vetorial F F F x 0 x_0 x 0
F : R 3 → R 3 F ∈ C 1 x 0 ∈ R 3 \begin{array}{lll}
F: \R^3 \to \R^3 & F \in C^1 & x_0 \in \R^3
\end{array} F : R 3 → R 3 F ∈ C 1 x 0 ∈ R 3 ∫ S ϵ rot F ⋅ n ⃗ = ∫ ∂ S ϵ F d g \int_{S_{\epsilon}} \rot F \cdot \vec n = \int_{\partial S_{\epsilon}} F \d g ∫ S ϵ rot F ⋅ n = ∫ ∂ S ϵ F d g ∫ S ϵ rot F ⋅ n ⃗ ≈ rot F ( x 0 ) ⋅ n ⃗ ∫ S ϵ 1 d S = rot F ( x 0 ) ⋅ n ⃗ a ˊ rea ( S ϵ ) \int_{S_{\epsilon}} \rot F \cdot \vec n \approx \rot F(x_0) \cdot \vec n \int_{S_{\epsilon}} 1 \d S = \rot F(x_0) \cdot \vec n\ \text{ área}(S_{\epsilon}) ∫ S ϵ rot F ⋅ n ≈ rot F ( x 0 ) ⋅ n ∫ S ϵ 1 d S = rot F ( x 0 ) ⋅ n a ˊ rea ( S ϵ ) rot F ( x 0 ) ⋅ n ⃗ = lim ϵ → 0 1 a ˊ rea ( S ϵ ) ∫ ∂ S ϵ F d g \rot F(x_0) \cdot \vec n = \lim_{\epsilon \to 0} \frac{1}{\text{área}(S_{\epsilon})} \int_{\partial S_{\epsilon}} F \d g rot F ( x 0 ) ⋅ n = ϵ → 0 lim a ˊ rea ( S ϵ ) 1 ∫ ∂ S ϵ F d g rot F ( x 0 ) ⋅ n ⃗ \rot F(x_0) \cdot \vec n rot F ( x 0 ) ⋅ n F F F n ⃗ \vec n n rot F ( x 0 ) \rot F(x_0) rot F ( x 0 )
A intensidade do trabalho de F F F
rot F ( x 0 ) ⋅ rot F ( x 0 ) ∣ ∣ rot F ( x 0 ) ∣ ∣ = ∣ ∣ rot F ( x 0 ) ∣ ∣ 2 ∣ ∣ rot F ( x 0 ) ∣ ∣ = ∣ ∣ rot F ( x 0 ) ∣ ∣ \rot F(x_0) \cdot \frac{\rot F(x_0)}{|| \rot F(x_0)||} = \frac{||\rot F(x_0)||^2}{|| \rot F(x_0)||} = ||\rot F(x_0)|| rot F ( x 0 ) ⋅ ∣∣ rot F ( x 0 ) ∣∣ rot F ( x 0 ) = ∣∣ rot F ( x 0 ) ∣∣ ∣∣ rot F ( x 0 ) ∣ ∣ 2 = ∣∣ rot F ( x 0 ) ∣∣ Slides:
