Section 6.4 Divergence 

6.4 Divergence 

This section, we learn the divergence of a vector field. The motivation for this is flux of a vector field along a curve or a surface. First, we introduce the flux of a vector field along a curve. 

Definition: 

The flux of [latex]\mathbf{F}[/latex] across [latex]C[/latex] is line integral

\[\int_{C}\overrightarrow{F}\cdot\overrightarrow{N(t)}ds\]

where [latex]\overrightarrow{N(t)}[/latex] is the unit normal vector of the curve.

Theorem:Calculating Flux across a Curve 

Let [latex]\mathbf{F}=\lt P,Q\gt[/latex] be a vector field over [latex]\mathbb{R}^{2}[/latex] and let [latex]C[/latex] be a smooth curve with parameterization [latex]\overrightarrow{r(t)}=\lt x(t),y(t)\gt[/latex], [latex]a\le t\le b.[/latex] Let [latex]\overrightarrow{n(t)}=\lt y'(t),-x'(t)\gt[/latex] . The flux of [latex]\mathbf{F}[/latex] across [latex]C[/latex] is 

\[\int_{C}\overrightarrow{F(r(t))}\cdot\overrightarrow{n(t)}dt=\int_{a}^{b}P(\overrightarrow{r(t)})y'(t)dt-Q(\overrightarrow{r(t)})x'(t)dt=\int_{C}Pdy-Qdx\]

Example 1: Calculate the flux of [latex]\overrightarrow{F}=\lt 2x,2y\gt[/latex] across a unit circle oriented counterclockwise.

Exercise 1: Calculate the flux of [latex]\overrightarrow{F}=\lt y,-x\gt[/latex] across a unit circle oriented counterclockwise.

Computing the integration over a curve may not be easy. Hence we want an alternative way for the computation. We introduce the divergence

of a vector field. 

Definition:

If [latex]\mathbf{F}=\lt P,Q,R\gt[/latex] or [latex]\mathbf{F}=\lt P,Q\gt[/latex] is a vector field in [latex]\mathbb{R}^{3}[/latex] or [latex]\mathbb{R}^{2}[/latex] and [latex]P_{x},Q_{y}[/latex] and [latex]R_{z}[/latex] all exist, then the divergence of [latex]\mathbf{F}[/latex] is defined

by  \[ \text{div}\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\text{ for }\mathbb{R}^{3}\]

or 

\[\text{div}\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\text{ for }\mathbb{R}^{2}.\]

Example 2: Find the divergence of [latex]\mathbf{F}(x,y,z)=xye^{z}\overrightarrow{i}+yze^{x}\overrightarrow{k}.[/latex]

Exercise 2: Find the divergence of [latex]\mathbf{F}(x,y,z)=xe^{y}\overrightarrow{j}+ze^{y}\overrightarrow{k}.[/latex]

Example 3: Find the divergence of [latex]\mathbf{F}(x,y,z)=\lt e^{x}\text{sin}(y),e^{y}\text{sin}(z),e^{z}\text{sin}(x)\gt .[/latex]

Exercise 3: Find the divergence of [latex]\mathbf{F}(x,y,z)=\lt e^{y}\text{cos}(z),e^{x}\text{cos}(y),e^{z}\text{cos}(x)\gt .[/latex]

Theorem:

If [latex]\mathbf{F}=\lt P,Q,R\gt[/latex] is a vector field on [latex]\mathbb{R}^{3}[/latex], and [latex]P,Q[/latex] and [latex]R[/latex] have continuous second order paritial derivatives then

\[\text{div}(\text{curl}(\mathbf{F}))=0.\]

Example 4: Is there a vector filed such that [latex]\text{curl}(\mathbf{F}(x,y,z))=\lt x\text{sin}(y),\text{cos}(y),z-xy\gt[/latex]? Explain.

Exercise 4: Is there a vector filed such that [latex]\text{curl}([/latex][latex]\mathbf{F}(x,y,z))=\lt e^{y}\text{cos}(z),e^{x}\text{cos}(y),e^{z}\text{cos}(x)\gt[/latex]? Explain.

Theorem: Calculating Flux across a Curve using Green’s Theorem 

Let [latex]\mathbf{F}=\lt P,Q\gt[/latex] be a vector field over [latex]\mathbb{R}^{2}[/latex], The flux of [latex]\mathbf{F}[/latex] across [latex]C[/latex] that enclose a region [latex]D[/latex] is

\[\int_{C}\overrightarrow{F(r(t))}\cdot\overrightarrow{n(t)}dt=\int\int_{D}\text{div}\mathbf{F}dA\]

Example 5: Calculate the flux of [latex]\overrightarrow{F}=\lt x^{2}+\text{cos}(y),\text{tan}(x)+3y\gt[/latex] along a curve [latex]C[/latex] from [latex](0,0)[/latex] to [latex](1,0)[/latex] to [latex](1,1)[/latex] to [latex](0,1)[/latex] and back to [latex](0,0)[/latex] with the positive orientation. 

Exercise 5: Calculate the flux of [latex]\overrightarrow{F}=\lt e^{x}+\text{sin}^{-1}(y),e^{y}+\text{sec}(x)\gt[/latex] along a curve [latex]C[/latex] from [latex](1,1)[/latex] to [latex](2,1)[/latex] to [latex](2,2)[/latex] to [latex](1,2)[/latex] and back to [latex](1,1)[/latex] with the positive orientation. 

Example 6: Calculate the flux of [latex]\overrightarrow{F}=\lt x^{2}+y^{2},y^{2}-x^{2}\gt[/latex] along a curve [latex]C[/latex] from [latex](1,0)[/latex] to [latex](1,-1)[/latex] to [latex](0,0)[/latex] and back to [latex](1,0)[/latex].

Exercise 6: Calculate the flux of [latex]\overrightarrow{F}=\lt x^{3}-y^{2},x^{2}-y^{2}\gt[/latex] along a curve [latex]C[/latex] from [latex](0,0)[/latex] to [latex](1,0)[/latex] to [latex](1,1)[/latex] and back to [latex](0,0)[/latex].

If [latex]\mathbf{F}(x,y,z)[/latex] is the velocity of a fluid (or gas), then [latex]\text{div}\mathbf{F}(x,y,z)[/latex] represents the net rate of change (with respect to time) of the mass of fluid (or gas) flowing from the point [latex](x,y,z)[/latex] per unit volume. In other words, [latex]\text{div}\mathbf{F}(x,y,z)[/latex] measures the tendency of the fluid to diverge from the point [latex](x,y,z)[/latex]. If [latex]\text{div}\mathbf{F}=0[/latex], then [latex]\mathbf{F}[/latex] is said to be incompressible.

Group work:

1. Find the divergence of [latex]\mathbf{F}[/latex] at a given point [latex]P[/latex].

a. [latex]\mathbf{F}=\lt xy^{2}z^{4},2x^{2}y+z,y^{3}z^{2}\gt[/latex] and [latex]P=(1,1,-1)[/latex].

b. [latex]\mathbf{F}=e^{x}\text{sin}(y)\overrightarrow{i}-e^{x}\text{cos}(y)\overrightarrow{j}[/latex] and [latex]P=(0,0,2)[/latex]. 

c. [latex]\mathbf{F}=\lt e^{-xy},e^{xz},e^{yz}\gt[/latex] and [latex]P=(3,2,0)[/latex]. 

2. Calculate the flux of [latex]\overrightarrow{F}=\lt x^{3}-y^{3},y^{3}+x^{3}\gt[/latex] along an ellipse [latex]C:x^{2}+9y^{2}=9[/latex] with positive orientation.

3. Calculate the work done by [latex]\overrightarrow{F}=\lt x^{3}-y^{3},y^{3}+x^{3}\gt[/latex] by moving an object along an ellipse [latex]C:x^{2}+9y^{2}=9[/latex] with positive orientation.