Flux in 2D (Part 2)

3. Outward flux across a closed plane curve

 Recall that: If $C$ is a piecewise-smooth, simple closed curve, the net outward flux of a vector field $\mathbf v =v_1(x,y)\,\mathbf i +v_2(x,y)\,\mathbf j$ across $C$ is given by $$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n\, dS$$ where $\mathbf n$ is a unit vector normal to $C$, directed outward from the region bounded by $C$.

Example 1. Suppose that $C$ is a circle of radius $r$, centre the origin. Let $\mathbf v=x\,\mathbf i+y\,\mathbf j$. In this case, the unit normal vector to $C$ is in the same direction as $\mathbf v$. Thus
$$\mathbf v \cdot \mathbf n =|\mathbf v|=r.$$
It follows that
$$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n dS=\oint_C r\,dS=r\oint_C dS=2\pi r^2.$$

The following applet shows a dynamic view of Example 1. Change the radius of the circle to check the general solution.

Example 2. Consider the same circle $C$ of radius $r$, centre the origin. But now we have $\mathbf v=-y\,\mathbf i+x\,\mathbf j$. In this case, $\mathbf v$ is perpendicular to the unit normal vector $\mathbf n$. Thus
$$\mathbf v \cdot \mathbf n =0.$$
Therefore
$$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n\, dS=0.$$

The following applet shows a dynamic view of Example 2. Change the radius of the circle to check the general solution.

Observation: The previous examples make sense physically. In the Example 1, the fluid is spewing out the origin, and thus flowing through $C$. On the other hand, in Example 2, the fluid is spinning around the origin. None of it crosses $C$.

 Flux form of Green's theorem If $C$ is a positively oriented simple closed curve enclosing a region $D$ and $\mathbf v =v_1(x,y)\,\mathbf i +v_2(x,y)\,\mathbf j$ then $$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n \,dS=\iint_D\text{div}(\mathbf v )\,dA$$ where $\text{div} (\mathbf v)=\dfrac{\partial v_1}{\partial x}+\dfrac{\partial v_2}{\partial y}$.

Example 3. Consider again the example of a circle $C$ of radius $r$, centre at the origin, and the velocity vector $\mathbf v =x\, \mathbf i +y\, \mathbf j$. Since
$$\text{div}(\mathbf v)=1+1=2,$$
then
$$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n \,dS=\iint_D\text{div}(\mathbf v )\,dA=\iint_D 2\,dA=2\pi r^2.$$

If we move the circle away from the origin, the flux is still $2\pi r^2$. However, computing $\oint_C \mathbf v \cdot \mathbf n \,dS$ becomes quite hard.

The following applet shows a dynamic representation of Example 3. Move the circle around. Observe that the flux remains constant. You can also observe the flux for a semicircle by dragging the slider for the partial path.

Key Concept

In the context of fluids the divergence measures how much fluid is being added (or taken away); these are known as sources (or sinks). For $\mathbf v =x\, \mathbf i +y\, \mathbf j$, fluid is being added everywhere (imagine rain falling on the ground and then flowing from the origin). For $\mathbf v =-y\, \mathbf i +x\, \mathbf j$ the divergence is zero. No fluid is being added or removed, there are no sources or sinks.

More

Cálculo Diferencial: Método para encontrar la velocidad de un movimiento cuando se conoce la distancia recorrida en un tiempo dado.
Cálculo Integral: Método para encontrar la distancia recorrida cuando se conoce la velocidad.
Intuitivamente:
Para encontrar la distancia recorrida de un objeto, cuando se conoce la velocidad, se recurre al cálculo integral, es decir, se debe calcular el área bajo la curva que representa la dependencia de la velocidad respecto del tiempo.
Para encontrar la velocidad de un movimiento cuando se conoce la distancia recorrida en un tiempo dado, se recurre al cálculo diferencial, es decir, se debe calcular la derivada de la curva que representa la dependencia de la distancia respecto del tiempo.
El problema de la integración es recíproco al problema de derivación y viceversa.
- Al integrar, función velocidad, se calcula distancia. - Al derivar, función distancia, se calcula velocidad
Applets de Geogebra
Representación del movimiento. En estos applets pueden modificar…

Representaciones en 3D: Espiral y curva paramétrica de pi

Otro uso de proyecciones ortográficas con Geogebra.

1. Curva paramétrica para representar a $\pi$
Para generar la curva que representa a $\pi$ se requiere utilizar una ecuación paramétrica.

En matemáticas, una ecuación paramétrica permite representar una o varias curvas o superficies en el plano o en el espacio, mediante valores arbitrarios o mediante una constante, llamada parámetro, en lugar de mediante una variable independiente de cuyos valores se desprenden los de la variable dependiente.
Por ejemplo: Dada la ecuación $y = x^2$, una parametrización tendrá la forma $$\begin{cases} x = u (t) \\ y = v (t) \end{cases}$$
Una parametrización posible sería $$\begin{cases} x = t \\ y = t^2 \end{cases}$$
Una circunferencia con centro en el origen de coordenadas y radio $r$ verifica que $x^2 + y^2 =r^2$.
Una expresión paramétrica de la circunferencia es $\begin{cases} x = r \cos t \\ y = r \sin t \end{cases}$
1.1 Curva $\pi$:
En nuestro caso, para generar la curva $\pi$, es necesario defini…

Möbius transformations with stereographic projections

A Möbius transformation of the plane is a rational function of the form $$f(z) = \frac{a z + b}{c z + d}$$ of one complex variable $z$. Here the coefficients $a, b, c, d$ are complex numbers satisfying $ad - bc\neq 0.$
Geometrically, a Möbius transformation can be obtained by stereographic projection of the complex plane onto an admissible sphere in $\mathbb R^3$, followed by a rigid motion of the sphere in $\mathbb R^3$ which maps it to another admissible sphere, followed by stereographic projection back to the plane.
Inversion

A Möbius transformation is a combination of dilatation, inversion, translation, and rotation.
The following applet shows the stereographic projection representing different Möbius transformations. Move the sliders to see what happens.

Made with GeoGebra, link here: http://tube.geogebra.org/student/m839839. This applet was made based on the work of D. N. Arnold and J. Rogness.