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 $C$.
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.
$$\mathbf v \cdot \mathbf n =|\mathbf v|=r.$$
It follows that
$$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n dS=\oint_C r\,dS=r\oint_C dS=2\pi r^2.$$
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 $\text{div} (\mathbf v)=\dfrac{\partial v_1}{\partial x}+\dfrac{\partial v_2}{\partial y}$. |
$$\text{div}(\mathbf v)=1+1=2,$$
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.
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.