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some typos.
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Fang Liu
2023-07-20 09:38:44 +08:00
parent 8fb5beb552
commit 7d56001593
5 changed files with 78 additions and 78 deletions
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72
quarto/integral_vector_calculus/stokes_theorem.qmd
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@@ -118,7 +118,7 @@ The total area under the blue curve from $a$ to $b$, is found by adding the area
Let's consider now what an integral over the boundary would mean. The region, or interval, $[x_{i-1}, x_i]$ has a boundary that clearly consists of the two points $x_{i-1}$ and $x_i$. If we *orient* the boundary, as we need to for higher dimensional boundaries, using the outward facing direction, then the oriented boundary at the right-hand end point, $x_i$, would point towards $+\infty$ and the left-hand end point, $x_{i-1}$, would be oriented to point to $-\infty$. An "integral" on the boundary of $F$ would naturally be $F(b) \times 1$ plus $F(a) \times -1$, or $F(b)-F(a)$.
With this choice of integral over the boundary, we can see much cancellation arises were we to compute this integral for each piece, as we would have with $a=x_0 < x_1 < \cdots x_{n-1} < x_n=b$:
With this choice of integral over the boundary, we can see much cancellation arises were we to compute this integral for each piece, as we would have with $a=x_0 < x_1 < \cdots < x_{n-1} < x_n=b$:
$$
@@ -199,7 +199,7 @@ $$
\right) \Delta{x}\Delta{y} .
$$
We interpret the right hand side as a Riemann sum approximation for the $2$ dimensional integral of the function $f(x,y) = \frac{\partial{F_x}}{\partial{y}} - \frac{\partial{F_y}}{\partial{x}}=\text{curl}(F)$, the two-dimensional curl. Were the green squares continued to fill out the large blue square, then the sum of these terms would approximate the integral
We interpret the right hand side as a Riemann sum approximation for the $2$ dimensional integral of the function $f(x,y) = \frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}}=\text{curl}(F)$, the two-dimensional curl. Were the green squares continued to fill out the large blue square, then the sum of these terms would approximate the integral
$$
@@ -375,7 +375,7 @@ Fx, Fy = F(x,y)
diff(Fy, x) - diff(Fx, y) |> simplify
```
As the integrand is $00$, $\iint_D \left( \partial{F_y}/{\partial{x}}-\partial{F_xy}/{\partial{y}}\right)dA = 0$, as well. But,
As the integrand is $0$, $\iint_D \left( \partial{F_y}/{\partial{x}}-\partial{F_xy}/{\partial{y}}\right)dA = 0$, as well. But,
$$
@@ -394,7 +394,7 @@ That is, for this example, Green's theorem does **not** apply, as the two integr
A simple closed curve is one that does not cross itself. Green's theorem applies to regions bounded by curves which have finitely many crosses provided the orientation used is consistent throughout.
Consider the curve $y = f(x)$, $a \leq x \leq b$, assuming $f$ is continuous, $f(a) > 0$, and $f(b) < 0$. We can use Green's theorem to compute the signed "area" under under $f$ if we consider the curve in $R^2$ from $(b,0)$ to $(a,0)$ to $(a, f(a))$, to $(b, f(b))$ and back to $(b,0)$ in that orientation. This will cross at each zero of $f$.
Consider the curve $y = f(x)$, $a \leq x \leq b$, assuming $f$ is continuous, $f(a) > 0$, and $f(b) < 0$. We can use Green's theorem to compute the signed "area" under $f$ if we consider the curve in $R^2$ from $(a,0)$ to $(b,0)$ to $(b, f(b))$, to $(a, f(a))$ and back to $(a,0)$ in that orientation. This will cross at each zero of $f$.
```{julia}
@@ -403,11 +403,11 @@ Consider the curve $y = f(x)$, $a \leq x \leq b$, assuming $f$ is continuous, $f
a, b = pi/2, 3pi/2
f(x) = sin(x)
p = plot(f, a, b, legend=false, xticks=nothing, border=:none, color=:green)
arrow!(p, [3pi/4, f(3pi/4)], 0.01*[1,cos(3pi/4)], color = :green)
arrow!(p, [5pi/4, f(5pi/4)], 0.01*[1,cos(5pi/4)], color = :green)
arrow!(p, [a,0], [0, f(a)], color=:red)
arrow!(p, [b, f(b)], [0, -f(b)], color=:blue)
arrow!(p, [b, 0], [a-b, 0], color=:black)
arrow!(p, [3pi/4, f(3pi/4)], -0.01*[1,cos(3pi/4)], color = :green)
arrow!(p, [5pi/4, f(5pi/4)], -0.01*[1,cos(5pi/4)], color = :green)
arrow!(p, [a,f(a)], [0, -f(a)], color=:red)
arrow!(p, [b, 0], [0, f(b)], color=:blue)
arrow!(p, [a, 0], [b-a, 0], color=:black)
del = -0.1
annotate!(p, [(a,del, "a"), (b,-del,"b")])
p
@@ -418,8 +418,8 @@ Let $A$ label the red line, $B$ the green curve, $C$ the blue line, and $D$ the
\begin{align*}
\int_A (xdy - ydx) &= a f(a)\\
\int_C (xdy - ydx) &= b(-f(b))\\
\int_A (xdy - ydx) &= a(-f(a))\\
\int_C (xdy - ydx) &= b f(b)\\
\int_D (xdy - ydx) &= 0\\
\end{align*}
@@ -430,7 +430,7 @@ Finally the integral over $B$, using integration by parts:
\begin{align*}
\int_B F(\vec{r}(t))\cdot \frac{d\vec{r}(t)}{dt} dt &=
\int_b^a \langle -f(t),t)\rangle\cdot\langle 1, f'(t)\rangle dt\\
\int_b^a \langle -f(t),t \rangle\cdot\langle 1, f'(t)\rangle dt\\
&= \int_a^b f(t)dt - \int_a^b tf'(t)dt\\
&= \int_a^b f(t)dt - \left(tf(t)\mid_a^b - \int_a^b f(t) dt\right).
\end{align*}
@@ -489,7 +489,7 @@ Adding the two gives $4\pi - \pi = \pi \cdot(b^2 - a^2)$, with $b=2$ and $a=1$.
#### Flow not flux
Green's theorem has a complement in terms of flow across $C$. As $C$ is positively oriented (so the bounded interior piece is on the left of $\hat{T}$ as the curve is traced), a normal comes by rotating $90^\circ$ counterclockwise. That is if $\hat{T} = \langle a, b\rangle$, then $\hat{N} = \langle b, -a\rangle$.
Green's theorem has a complement in terms of flow across $C$. As $C$ is positively oriented (so the bounded interior piece is on the left of $\hat{T}$ as the curve is traced), a normal comes by rotating $90^\circ$ clockwise. That is if $\hat{T} = \langle a, b\rangle$, then $\hat{N} = \langle b, -a\rangle$.
Let $F = \langle F_x, F_y \rangle$ and $G = \langle F_y, -F_x \rangle$, then $G\cdot\hat{T} = -F\cdot\hat{N}$. The curl formula applied to $G$ becomes
@@ -497,7 +497,7 @@ Let $F = \langle F_x, F_y \rangle$ and $G = \langle F_y, -F_x \rangle$, then $G\
$$
\frac{\partial{G_y}}{\partial{x}} - \frac{\partial{G_x}}{\partial{y}} =
\frac{\partial{-F_x}}{\partial{x}}-\frac{\partial{(F_y)}}{\partial{y}}
\frac{\partial{(-F_x)}}{\partial{x}}-\frac{\partial{(F_y)}}{\partial{y}}
=
-\left(\frac{\partial{F_x}}{\partial{x}} + \frac{\partial{F_y}}{\partial{y}}\right)=
-\nabla\cdot{F}.
@@ -596,19 +596,19 @@ p
Again, the microscopic boundary integrals when added will give a macroscopic boundary integral due to cancellations.
But, as seen in the derivation of the divergence, only modified for $2$ dimensions, we have $\nabla\cdot{F} = \lim \frac{1}{\Delta S} \oint_C F\cdot\hat{N}$, so for each cell
But, as seen in the derivation of the divergence, only modified for $2$ dimensions, we have $\nabla\cdot{F} = \lim \frac{1}{\Delta S} \oint_C F\cdot\hat{N} ds$, so for each cell
$$
\oint_{C_i} F\cdot\hat{N} \approx \left(\nabla\cdot{F}\right)\Delta{x}\Delta{y},
\oint_{C_i} F\cdot\hat{N} ds \approx \left(\nabla\cdot{F}\right)\Delta{x}\Delta{y},
$$
an approximating Riemann sum for $\iint_D \nabla\cdot{F} dA$. This yields:
$$
\oint_C (F \cdot\hat{N}) dA =
\sum_i \oint_{C_i} (F \cdot\hat{N}) dA \approx
\oint_C (F \cdot\hat{N}) ds =
\sum_i \oint_{C_i} (F \cdot\hat{N}) ds \approx
\sum \left(\nabla\cdot{F}\right)\Delta{x}\Delta{y} \approx
\iint_S \nabla\cdot{F}dA,
$$
@@ -630,11 +630,11 @@ The integral of the flow across $C$ consists of $4$ parts. By symmetry, they all
$$
\int_C F \cdot \hat{N} ds=
\int_{-1}^1 \langle F_x, F_y\rangle\cdot\langle 0, 1\rangle ds =
\int_{-1}^1 b dy = 2b.
\int_{-1}^1 \langle F_x, F_y\rangle\cdot\langle 1, 0\rangle ds =
\int_{-1}^1 a dy = 2a.
$$
Integrating across the top will give $2a$, along the bottom $2a$, and along the left side $2b$ totaling $4(a+b)$.
Integrating across the top will give $2b$, along the bottom $2b$, and along the left side $2a$ totaling $4(a+b)$.
---
@@ -822,7 +822,7 @@ $$
\oint_C B\cdot\hat{T} ds = \mu_0 I.
$$
The goal here is to re-express this integral law to produce a law at each point of the field. Let $S$ be a surface with boundary $C$, Let $J$ be the current density - $J=\rho v$, with $\rho$ the density of the current (not time-varying) and $v$ the velocity. The current can be re-expressed as $I = \iint_S J\cdot\hat{n}dA$. (If the current flows through a wire and $S$ is much bigger than the wire, this is still valid as $\rho=0$ outside of the wire.)
The goal here is to re-express this integral law to produce a law at each point of the field. Let $S$ be a surface with boundary $C$, Let $J$ be the current density - $J=\rho v$, with $\rho$ the density of the current (not time-varying) and $v$ the velocity. The current can be re-expressed as $I = \iint_S J\cdot\hat{N}dA$. (If the current flows through a wire and $S$ is much bigger than the wire, this is still valid as $\rho=0$ outside of the wire.)
We then have:
@@ -859,7 +859,7 @@ $$
-\iint_S \left(\frac{\partial{B}}{\partial{t}}\cdot\hat{N}\right)dS =
-\frac{\partial{\phi}}{\partial{t}} =
\oint_C E\cdot\hat{T}ds =
\iint_S (\nabla\times{E}) dS.
\iint_S (\nabla\times{E})\cdot\hat{N} dS.
$$
This is true for any capping surface for $C$. Shrinking $C$ to a point means it will hold for each point in $R^3$. That is:
@@ -884,10 +884,10 @@ Green's theorem gave a characterization of $2$-dimensional conservative fields,
Stokes's theorem can be used to show the first and fourth are equivalent.
First, if $0 = \oint_C F\cdot\hat{T} ds$, then by Stokes' theorem $0 = \int_S \nabla\times{F} dS$ for any orientable surface $S$ with boundary $C$. For a given point, letting $C$ shrink to that point can be used to see that the cross product must be $0$ at that point.
First, if $0 = \oint_C F\cdot\hat{T} ds$, then by Stokes' theorem $0 = \iint_S \nabla\times{F}\cdot\hat{N} dS$ for any orientable surface $S$ with boundary $C$. For a given point, letting $C$ shrink to that point can be used to see that the cross product must be $0$ at that point.
Conversely, if the cross product is zero in a simply connected region, then take any simple closed curve, $C$ in the region. If the region is [simply connected](http://math.mit.edu/~jorloff/suppnotes/suppnotes02/v14.pdf) then there exists an orientable surface, $S$ in the region with boundary $C$ for which: $\oint_C F\cdot{N} ds = \iint_S (\nabla\times{F})\cdot\hat{N}dS= \iint_S \vec{0}\cdot\hat{N}dS = 0$.
Conversely, if the cross product is zero in a simply connected region, then take any simple closed curve, $C$ in the region. If the region is [simply connected](http://math.mit.edu/~jorloff/suppnotes/suppnotes02/v14.pdf) then there exists an orientable surface, $S$ in the region with boundary $C$ for which: $\oint_C F\cdot\hat{T} ds = \iint_S (\nabla\times{F})\cdot\hat{N}dS= \iint_S \vec{0}\cdot\hat{N}dS = 0$.
The construction of a scalar potential function from the field can be done as illustrated in this next example.
@@ -970,7 +970,7 @@ This is also easy, as `Ft` has only an `x` component and `rp` has only `y` and `
In two dimensions the vector field $F(x,y) = \langle -y, x\rangle/(x^2+y^2) = S(x,y)/\|R\|^2$ is irrotational ($0$ curl) and has $0$ divergence, but is *not* conservative in $R^2$, as with $C$ being the unit disk we have $\oint_C F\cdot\hat{T}ds = \int_0^{2\pi} \langle -\sin(\theta),\cos(\theta)\rangle \cdot \langle-\sin(\theta), \cos(\theta)\rangle/1 d\theta = 2\pi$. This is because $F$ is not continuously differentiable at the origin, so the path $C$ is not in a simply connected domain where $F$ is continuously differentiable. (Were $C$ to avoid the origin, the integral would be $0$.)
In three dimensions, removing a single point in a domain does change simple connectedness, but removing an entire line will. So the function $F(x,y,z) =\langle -y,x,0\rangle/(x^2+y^2)\rangle$ will have $0$ curl, $0$ divergence, but won't be conservative in a domain that includes the $z$ axis.
In three dimensions, removing a single point in a domain does not change simple connectedness, but removing an entire line will. So the function $F(x,y,z) =\langle -y,x,0\rangle/(x^2+y^2)\rangle$ will have $0$ curl, $0$ divergence, but won't be conservative in a domain that includes the $z$ axis.
However, the function $F(x,y,z) = \langle x, y,z\rangle/\sqrt{x^2+y^2+z^2}$ has curl $0$, except at the origin. However, $R^3$ less the origin, as a domain, is simply connected, so $F$ will be conservative.
@@ -983,15 +983,15 @@ The divergence theorem is a consequence of a simple observation. Consider two ad
$$
\oint_S F\cdot{N} dA = \sum \oint_{S_i} F\cdot{N} dA.
\oint_S F\cdot\hat{N} dA = \sum \oint_{S_i} F\cdot\hat{N} dA.
$$
If the partition provides a microscopic perspective, then the divergence approximation $\nabla\cdot{F} \approx (1/\Delta{V_i}) \oint_{S_i} F\cdot{N} dA$ can be used to say:
If the partition provides a microscopic perspective, then the divergence approximation $\nabla\cdot{F} \approx (1/\Delta{V_i}) \oint_{S_i} F\cdot\hat{N} dA$ can be used to say:
$$
\oint_S F\cdot{N} dA =
\sum \oint_{S_i} F\cdot{N} dA \approx
\oint_S F\cdot\hat{N} dA =
\sum \oint_{S_i} F\cdot\hat{N} dA \approx
\sum (\nabla\cdot{F})\Delta{V_i} \approx
\iiint_V \nabla\cdot{F} dV,
$$
@@ -1046,7 +1046,7 @@ In fact, all $6$ sides will be $0$, as in this case $F \cdot \hat{i} = xy$ and a
As such, the two sides of the Divergence theorem are both $0$, so the theorem is verified.
###### Example
##### Example
(From Strang) If the temperature inside the sun is $T = \log(1/\rho)$ find the *heat* flow $F=-\nabla{T}$; the source, $\nabla\cdot{F}$; and the flux, $\iint F\cdot\hat{N}dS$. Model the sun as a ball of radius $\rho_0$.
@@ -1193,7 +1193,7 @@ The simplification done by SymPy masks the presence of $R^{-5/2}$ when taking th
$$
0 = \iiint_V \nabla\cdot{F} dV =
\oint_S F\cdot{N}dS = \oint_S \frac{R}{\|R\|^3} \cdot{R} dS =
\oint_S F\cdot\hat{N}dS = \oint_S \frac{R}{\|R\|^3} \cdot{R} dS =
\oint_S 1 dS = 4\pi.
$$
@@ -1249,7 +1249,7 @@ numericq(val)
Let $\hat{N} = \langle \cos(t), \sin(t) \rangle$ and $\hat{T} = \langle -\sin(t), \cos(t)\rangle$. Then polar coordinates can be viewed as the parametric curve $\vec{r}(t) = r(t) \hat{N}$.
Applying Green's theorem to the vector field $F = \langle -y, x\rangle$ which along the curve is $r(t) \hat{T}$ we know the area formula $(1/2) (\int xdy - \int y dx)$. What is this in polar coordinates (using $\theta=t$?) (Using $(r\hat{N}' = r'\hat{N} + r \hat{N}' = r'\hat{N} +r\hat{T}$ is useful.)
Applying Green's theorem to the vector field $F = \langle -y, x\rangle$ which along the curve is $r(t) \hat{T}$ we know the area formula $(1/2) (\int xdy - \int y dx)$. What is this in polar coordinates (using $\theta=t$?) (Using $(r\hat{N})' = r'\hat{N} + r \hat{N}' = r'\hat{N} +r\hat{T}$ is useful.)
```{julia}
@@ -1338,7 +1338,7 @@ answ = 2
radioq(choices, answ, keep_order=true)
```
Along path $C$, $F(x,y) = [1,x]$ and $\hat{T}=-\hat{i}$ so $F\cdot\hat{T} = -1$. The path integral $\int_C (F\cdot\hat{T})ds = -1$. What is the value of the path integral over $A$?
Along path $C$, $F(x,y) = [1,x]$ and $\hat{T}=-\hat{i}$ so $F\cdot\hat{T} = -1$. The path integral $\int_C (F\cdot\hat{T})ds = -1$. What is the value of the path integral over $B$?
```{julia}
@@ -1441,7 +1441,7 @@ radioq(choices, answ, keep_order=true)
###### Question
Let $R(x,y,z) = \langle x, y, z\rangle$ and $\rho = \|R\|^2$. If $F = 2R/\rho^2$ then $F$ is the gradient of a potential. Which one?
Let $R(x,y,z) = \langle x, y, z\rangle$ and $\rho = \|R\|^2$. If $F = 2R/\rho$ then $F$ is the gradient of a potential. Which one?
```{julia}
@@ -1545,7 +1545,7 @@ The diagram emphasizes a few different things:
The one for the curl in $n=2$ is Green's theorem: $\iint_S \nabla\times{F}dA = \oint_{\partial{S}} F\cdot d\vec{r}$.
The one for the curl in $n=3$ is Stoke's theorem: $\iint S \nabla\times{F}dA = \oint_{\partial{S}} F\cdot d\vec{r}$. Finally, the divergence for $n=3$ is the divergence theorem $\iint_V \nabla\cdot{F} dV = \iint_{\partial{V}} F dS$.
The one for the curl in $n=3$ is Stoke's theorem: $\iint_S \nabla\times{F}dA = \oint_{\partial{S}} F\cdot d\vec{r}$. Finally, the divergence for $n=3$ is the divergence theorem $\iint_V \nabla\cdot{F} dV = \iint_{\partial{V}} F dS$.
* Working left to right along a row of the diagram, applying two steps of these operations yields: