make pdf file generation work

quarto/integral_vector_calculus/div_grad_curl.qmd

@@ -216,8 +216,8 @@ annotate!([
 Let $F=\langle F_x, F_y\rangle$. For small enough values of $\Delta{x}$ and $\Delta{y}$ the line integral, $\oint_C F\cdot d\vec{r}$ can be *approximated* by $4$ terms:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \left(F(x,y) \cdot \hat{i}\right)\Delta{x} &+
 \left(F(x+\Delta{x},y) \cdot \hat{j}\right)\Delta{y} +
 \left(F(x,y+\Delta{y}) \cdot (-\hat{i})\right)\Delta{x} +
@@ -228,8 +228,8 @@ F_x(x, y+\Delta{y}) (-\Delta{x}) + F_y(x,y) (-\Delta{y})\\
 &=
 (F_y(x + \Delta{x}, y) - F_y(x, y))\Delta{y} -
 (F_x(x, y+\Delta{y})-F_x(x,y))\Delta{x}.
-\end{align}
-$$
+\end{align*}
+
 
 The Riemann approximation allows a choice of evaluation point for Riemann integrable functions, and the choice here lends itself to further analysis. Were the above divided by $\Delta{x}\Delta{y}$, the area of the box, and a limit taken, partial derivatives appear to suggest this formula:
 
@@ -325,18 +325,18 @@ p
 Now we compute the *line integral*. Consider the top face, $S_1$, connecting $(x,y,z+\Delta z), (x + \Delta x, y, z + \Delta z),  (x + \Delta x, y + \Delta y, z + \Delta z),  (x, y + \Delta y, z + \Delta z)$, Using the *right hand rule*, parameterize the boundary curve, $C_1$, in a counter clockwise direction so the right hand rule yields the outward pointing normal ($\hat{k}$). Then the integral $\oint_{C_1} F\cdot \hat{T} ds$ is *approximated* by the following Riemann sum of $4$ terms:
 
 
-$$
+
 \begin{align*}
 F(x,y, z+\Delta{z}) \cdot \hat{i}\Delta{x}  &+ F(x+\Delta x, y, z+\Delta{z}) \cdot \hat{j}  \Delta y \\
 &+ F(x, y+\Delta y, z+\Delta{z}) \cdot (-\hat{i}) \Delta{x}  \\
 &+ F(x, y, z+\Delta{z}) \cdot (-\hat{j}) \Delta{y}.
 \end{align*}
-$$
+
 
 (The points $c_i$ are chosen from the endpoints of the line segments.)
 
 
-$$
+
 \begin{align*}
 \oint_{C_1} F\cdot \hat{T} ds
 &\approx  (F_y(x+\Delta x, y, z+\Delta{z})   \\
@@ -344,18 +344,18 @@ $$
 &-  (F_x(x,y + \Delta{y}, z+\Delta{z})  \\
 &- F_x(x, y, z+\Delta{z})) \Delta{x}
 \end{align*}
-$$
+
 
 As before, were this divided by the *area* of the surface, we have after rearranging and cancellation:
 
 
-$$
+
 \begin{align*}
 \frac{1}{\Delta{S_1}} \oint_{C_1} F \cdot \hat{T} ds &\approx
 \frac{F_y(x+\Delta x, y, z+\Delta{z}) - F_y(x, y, z+\Delta{z})}{\Delta{x}}\\
 &- \frac{F_x(x, y+\Delta y, z+\Delta{z}) - F_x(x, y, z+\Delta{z})}{\Delta{y}}.
 \end{align*}
-$$
+
 
 In the limit, as  $\Delta{S} \rightarrow 0$, this will converge to $\partial{F_y}/\partial{x}-\partial{F_x}/\partial{y}$.
 
@@ -366,7 +366,7 @@ Had the bottom of the box been used, a similar result would be found, up to a mi
 Unlike the two dimensional case, there are other directions to consider and here the other sides will yield different answers. Consider now the face connecting $(x,y,z), (x+\Delta{x}, y, z), (x+\Delta{x}, y, z + \Delta{z})$, and $ (x,y,z+\Delta{z})$ with outward pointing normal $-\hat{j}$. Let $S_2$ denote this face and $C_2$ describe its boundary. Orient this curve so that the right hand rule points in the $-\hat{j}$ direction (the outward pointing normal). Then, as before, we can approximate:
 
 
-$$
+
 \begin{align*}
 \oint_{C_2} F \cdot \hat{T} ds
 &\approx
@@ -377,7 +377,7 @@ F(x,y,z) \cdot \hat{i} \Delta{x} \\
 &= (F_z(x+\Delta{x},y,z) - F_z(x, y, z))\Delta{z} -
 (F_x(x,y,z+\Delta{z}) - F(x,y,z)) \Delta{x}.
 \end{align*}
-$$
+
 
 Dividing by $\Delta{S}=\Delta{x}\Delta{z}$ and taking a limit will give:
 
@@ -423,13 +423,12 @@ The curl has a formal representation in terms of a $3\times 3$ determinant, simi
 
 
 $$
-\text{curl}(F) = \det\left[
-\begin{array}{}
+\text{curl}(F) = \det
+\begin{bmatrix}
 \hat{i} & \hat{j} & \hat{k}\\
 \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\
 F_x & F_y & F_z
-\end{array}
-\right]
+\end{bmatrix}
 $$
 
 ---
@@ -474,7 +473,7 @@ The divergence, gradient, and curl all involve partial derivatives. There is a n
 This is a *vector differential operator* that acts on functions and vector fields through the typical notation to yield the three operations:
 
 
-$$
+
 \begin{align*}
 \nabla{f} &= \langle
 \frac{\partial{f}}{\partial{x}},
@@ -502,17 +501,17 @@ $$
 \frac{\partial}{\partial{y}},
 \frac{\partial}{\partial{z}}
 \rangle \times F =
-\det\left[
-\begin{array}{}
+\det
+\begin{bmatrix}
 \hat{i} & \hat{j} & \hat{k} \\
 \frac{\partial}{\partial{x}}&
 \frac{\partial}{\partial{y}}&
 \frac{\partial}{\partial{z}}\\
 F_x & F_y & F_z
-\end{array}
-\right],\quad\text{the curl}.
+\end{bmatrix}
+,\quad\text{the curl}.
 \end{align*}
-$$
+
 
 :::{.callout-note}
 ## Note
@@ -842,13 +841,13 @@ Let $f$ and $g$ denote scalar functions, $R^3 \rightarrow R$ and $F$ and $G$ be
 As with the sum rule of univariate derivatives, these operations satisfy:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \nabla(f + g) &= \nabla{f} + \nabla{g}\\
 \nabla\cdot(F+G) &= \nabla\cdot{F} + \nabla\cdot{G}\\
 \nabla\times(F+G) &= \nabla\times{F} + \nabla\times{G}.
-\end{align}
-$$
+\end{align*}
+
 
 ### Product rule
 
@@ -856,13 +855,13 @@ $$
 The product rule $(uv)' = u'v + uv'$ has related formulas:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \nabla{(fg)} &= (\nabla{f}) g + f\nabla{g} = g\nabla{f}  + f\nabla{g}\\
 \nabla\cdot{fF} &= (\nabla{f})\cdot{F} +  f(\nabla\cdot{F})\\
 \nabla\times{fF} &= (\nabla{f})\times{F} +  f(\nabla\times{F}).
-\end{align}
-$$
+\end{align*}
+
 
 ### Rules over cross products
 
@@ -870,13 +869,13 @@ $$
 The cross product of two vector fields is a vector field for which the divergence and curl may be taken. There are formulas to relate to the individual terms:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \nabla\cdot(F \times G) &= (\nabla\times{F})\cdot G - F \cdot (\nabla\times{G})\\
 \nabla\times(F \times G) &= F(\nabla\cdot{G}) - G(\nabla\cdot{F} + (G\cdot\nabla)F-(F\cdot\nabla)G\\
 &= \nabla\cdot(BA^t - AB^t).
-\end{align}
-$$
+\end{align*}
+
 
 The curl formula is more involved.
 
@@ -921,7 +920,7 @@ Second,
 This is not as clear, but can be seen algebraically as terms cancel. First:
 
 
-$$
+
 \begin{align*}
 \nabla\cdot(\nabla\times{F}) &=
 \langle
@@ -938,7 +937,7 @@ $$
 \left(\frac{\partial^2{F_x}}{\partial{z}\partial{y}} - \frac{\partial^2{F_z}}{\partial{x}\partial{y}}\right) +
 \left(\frac{\partial^2{F_y}}{\partial{x}\partial{z}} - \frac{\partial^2{F_x}}{\partial{y}\partial{z}}\right)
 \end{align*}
-$$
+
 
 Focusing on one component function, $F_z$ say, we see this contribution:
 
@@ -1014,7 +1013,7 @@ This is because of how the line integrals are oriented so that the right-hand ru
 The [invariance of charge](https://en.wikipedia.org/wiki/Maxwell%27s_equations#Charge_conservation) can be derived as a corollary of Maxwell's equation. The divergence of the curl of the magnetic field is $0$, leading to:
 
 
-```{mat}
+
 \begin{align*}
 0 &= \nabla\cdot(\nabla\times{B}) \\
 &=
@@ -1024,7 +1023,7 @@ The [invariance of charge](https://en.wikipedia.org/wiki/Maxwell%27s_equations#C
 &=
 \mu_0(\nabla\cdot{J} + \frac{\partial{\rho}}{\partial{t}}).
 \end{align*}
-```
+
 
 That is $\nabla\cdot{J} = -\partial{\rho}/\partial{t}$. This says any change in the charge density in time ($\partial{\rho}/\partial{t}$) is balanced off by a divergence in the electric current density ($\nabla\cdot{J}$). That is, charge can't be created or destroyed in an isolated system.
 
@@ -1048,25 +1047,25 @@ $$
 Without explaining why, these values can be computed using volume and surface integrals:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \phi(\vec{r}') &=
 \frac{1}{4\pi} \int_V \frac{\nabla \cdot F(\vec{r})}{\|\vec{r}'-\vec{r} \|} dV -
 \frac{1}{4\pi} \oint_S \frac{F(\vec{r})}{\|\vec{r}'-\vec{r} \|} \cdot \hat{N} dS\\
 A(\vec{r}') &= \frac{1}{4\pi} \int_V \frac{\nabla \times F(\vec{r})}{\|\vec{r}'-\vec{r} \|} dV +
 \frac{1}{4\pi} \oint_S \frac{F(\vec{r})}{\|\vec{r}'-\vec{r} \|} \times \hat{N} dS.
-\end{align}
-$$
+\end{align*}
+
 
 If $V = R^3$, an unbounded domain, *but* $F$ *vanishes* faster than $1/r$, then the theorem still holds with just the volume integrals:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \phi(\vec{r}') &=\frac{1}{4\pi} \int_V \frac{\nabla \cdot F(\vec{r})}{\|\vec{r}'-\vec{r} \|} dV\\
 A(\vec{r}') &= \frac{1}{4\pi} \int_V \frac{\nabla \times F(\vec{r})}{\|\vec{r}'-\vec{r}\|} dV.
-\end{align}
-$$
+\end{align*}
+
 
 ## Change of variable
 
@@ -1080,8 +1079,8 @@ Some details are [here](https://en.wikipedia.org/wiki/Curvilinear_coordinates),
 We restrict to $n=3$ and use $(x,y,z)$ for Cartesian coordinates and $(u,v,w)$ for an *orthogonal* curvilinear coordinate system, such as spherical or cylindrical. If $\vec{r} = \langle x,y,z\rangle$, then
 
 
-$$
-\begin{align}
+
+\begin{align*}
 d\vec{r} &= \langle dx,dy,dz \rangle = J \langle du,dv,dw\rangle\\
 &=
 \left[ \frac{\partial{\vec{r}}}{\partial{u}} \vdots
@@ -1090,8 +1089,8 @@ d\vec{r} &= \langle dx,dy,dz \rangle = J \langle du,dv,dw\rangle\\
 &= \frac{\partial{\vec{r}}}{\partial{u}} du +
 \frac{\partial{\vec{r}}}{\partial{v}} dv
 \frac{\partial{\vec{r}}}{\partial{w}} dw.
-\end{align}
-$$
+\end{align*}
+
 
 The term ${\partial{\vec{r}}}/{\partial{u}}$ is tangent to the curve formed by *assuming* $v$ and $w$ are constant and letting $u$ vary. Similarly for the other partial derivatives. Orthogonality assumes that at every point, these tangent vectors are orthogonal.
 
@@ -1138,7 +1137,7 @@ This uses orthogonality, so $\hat{e}_v \times \hat{e}_w$ is parallel to $\hat{e}
 The volume element is found by *projecting* $d\vec{r}$ onto the $\hat{e}_u$, $\hat{e}_v$, $\hat{e}_w$ coordinate system through $(d\vec{r} \cdot\hat{e}_u) \hat{e}_u$, $(d\vec{r} \cdot\hat{e}_v) \hat{e}_v$, and $(d\vec{r} \cdot\hat{e}_w) \hat{e}_w$. Then forming the triple scalar product to compute the volume of the parallelepiped:
 
 
-$$
+
 \begin{align*}
 \left[(d\vec{r} \cdot\hat{e}_u) \hat{e}_u\right] \cdot
 \left(
@@ -1149,7 +1148,7 @@ $$
 &=
 h_u h_v h_w  du dv dw,
 \end{align*}
-$$
+
 
 as the unit vectors are orthonormal, their triple scalar product is $1$ and $d\vec{r}\cdot\hat{e}_u = h_u du$, etc.
 
@@ -1214,21 +1213,21 @@ p
 The tangent vectors found from the partial derivatives of $\vec{r}$:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \frac{\partial{\vec{r}}}{\partial{r}} &=
 \langle \cos(\theta) \cdot \sin(\phi), \sin(\theta) \cdot \sin(\phi), \cos(\phi)\rangle,\\
 \frac{\partial{\vec{r}}}{\partial{\theta}} &=
 \langle -r\cdot\sin(\theta)\cdot\sin(\phi), r\cdot\cos(\theta)\cdot\sin(\phi), 0\rangle,\\
 \frac{\partial{\vec{r}}}{\partial{\phi}} &=
 \langle r\cdot\cos(\theta)\cdot\cos(\phi), r\cdot\sin(\theta)\cdot\cos(\phi), -r\cdot\sin(\phi) \rangle.
-\end{align}
-$$
+\end{align*}
+
 
 With this, we have $h_r=1$, $h_\theta=r\sin(\phi)$, and $h_\phi = r$. So that
 
 
-$$
+
 \begin{align*}
 dl &= \sqrt{dr^2 + (r\sin(\phi)d\theta^2) + (rd\phi)^2},\\
 dS_r &= r^2\sin(\phi)d\theta d\phi,\\
@@ -1236,7 +1235,7 @@ dS_\theta &= rdr d\phi,\\
 dS_\phi &= r\sin(\phi)dr d\theta, \quad\text{and}\\
 dV &= r^2\sin(\phi) drd\theta d\phi.
 \end{align*}
-$$
+
 
 The following visualizes the volume and the surface elements.
 
@@ -1292,7 +1291,7 @@ p
 If $f$ is a scalar function then $df = \nabla{f} \cdot d\vec{r}$ by the chain rule. Using the curvilinear coordinates:
 
 
-$$
+
 \begin{align*}
 df &=
 \frac{\partial{f}}{\partial{u}} du +
@@ -1303,7 +1302,7 @@ df &=
 \frac{1}{h_v}\frac{\partial{f}}{\partial{v}} h_vdv +
 \frac{1}{h_w}\frac{\partial{f}}{\partial{w}} h_wdw.
 \end{align*}
-$$
+
 
 But, as was used above, $d\vec{r} \cdot \hat{e}_u = h_u du$, etc. so $df$ can be re-expressed as:
 
@@ -1379,13 +1378,12 @@ where $S$ is a surface perpendicular to $\hat{N}$ with boundary $C$. For a small
 
 
 $$
-\nabla\times{F} = \det \left[
-\begin{array}{}
+\nabla\times{F} = \det
+\begin{bmatrix}
 h_u\hat{e}_u & h_v\hat{e}_v & h_w\hat{e}_w \\
 \frac{\partial}{\partial{u}} & \frac{\partial}{\partial{v}} & \frac{\partial}{\partial{w}} \\
 h_uF_u & h_v F_v & h_w F_w
-\end{array}
-\right].
+\end{bmatrix}.
 $$
 
 ---
@@ -1395,32 +1393,30 @@ For example, in cylindrical coordinates, the curl is:
 
 
 $$
-\det\left[
-\begin{array}{}
+\det
+\begin{bmatrix}
 \hat{r} & r\hat{\theta} & \hat{k} \\
 \frac{\partial}{\partial{r}} & \frac{\partial}{\partial{\theta}} & \frac{\partial}{\partial{z}} \\
 F_r & rF_\theta & F_z
-\end{array}
-\right]
+\end{bmatrix}
 $$
 
 Applying this to the function $F(r,\theta, z) = \hat{\theta}$ we get:
 
 
 $$
-\text{curl}(F) = \det\left[
-\begin{array}{}
+\text{curl}(F) = \det
+\begin{bmatrix}
 \hat{r} & r\hat{\theta} & \hat{k} \\
 \frac{\partial}{\partial{r}} & \frac{\partial}{\partial{\theta}} & \frac{\partial}{\partial{z}} \\
 0 & r & 0
-\end{array}
-\right] =
-\hat{k} \det\left[
-\begin{array}{}
+\end{bmatrix}
+=
+\hat{k} \det
+\begin{bmatrix}
 \frac{\partial}{\partial{r}} & \frac{\partial}{\partial{\theta}}\\
 0 & r
-\end{array}
-\right] =
+\end{bmatrix} =
 \hat{k}.
 $$
 
@@ -1646,13 +1642,12 @@ The curl then will then be
 
 
 $$
-\nabla\times{F} = \det \left[
-\begin{array}{}
+\nabla\times{F} = \det
+\begin{bmatrix}
 \hat{e}_r & r\sin\phi\hat{e}_\theta & r\hat{e}_\phi \\
 \frac{\partial}{\partial{r}} & \frac{\partial}{\partial{\theta}} & \frac{\partial}{\partial{phi}} \\
 F_r & r\sin\phi F_\theta & r F_\phi
-\end{array}
-\right].
+\end{bmatrix}.
 $$
 
 For a *radial* function $F = h(r)e_r$. (That is $F_r = h(r)$, $F_\theta=0$, and $F_\phi=0$. What is the curl of $F$?

quarto/integral_vector_calculus/double_triple_integrals.qmd

@@ -46,9 +46,12 @@ The multidimensional case will prove to be similar where a Riemann sum is used t
 #| echo: false
 imgfile = "figures/chrysler-building-in-new-york.jpg"
 caption = """How to estimate the volume contained within the Chrysler Building? One way might be to break the building up into tall vertical blocks based on its skyline;  compute the volume of each block using the formula of volume as area of the base times the height; and, finally, adding up the computed volumes This is the basic idea of finding volumes under surfaces using Riemann integration."""
-ImageFile(:integral_vector_calculus, imgfile, caption)
+#ImageFile(:integral_vector_calculus, imgfile, caption)
+nothing
 ```
 
+![How to estimate the volume contained within the Chrysler Building? One way might be to break the building up into tall vertical blocks based on its skyline;  compute the volume of each block using the formula of volume as area of the base times the height; and, finally, adding up the computed volumes This is the basic idea of finding volumes under surfaces using Riemann integration.](./figures/chrysler-building-in-new-york.jpg)
+
 ```{julia}
 #| hold: true
 #| echo: false
@@ -57,9 +60,14 @@ caption = """
 Computing the volume of a nano-block construction of the Chrysler building is easier than trying to find an actual tree at the Chrysler building, as we can easily compute the volume of columns of equal-sized blocks. Riemann sums are similar.
 """
 
-ImageFile(:integral_vector_calculus, imgfile, caption)
+#ImageFile(:integral_vector_calculus, imgfile, caption)
+nothing
 ```
 
+![
+Computing the volume of a nano-block construction of the Chrysler building is easier than trying to find an actual tree at the Chrysler building, as we can easily compute the volume of columns of equal-sized blocks. Riemann sums are similar.
+](./figures/chrysler-nano-block.png)
+
 The definition of the  multi-dimensional integral is more involved then the one-dimensional case due to the possibly increased complexity of the region. This will require additional [steps](https://math.okstate.edu/people/lebl/osu4153-s16/chapter10-ver1.pdf). The basic approach is as follows.
 
 
@@ -410,9 +418,12 @@ In [Ferzola](https://doi.org/10.2307/2687130) we can read a summary of Euler's t
 imgfile ="figures/strang-slicing.png"
 caption = L"""Figure 14.2 of Strang illustrating the slice when either $x$ is fixed or $y$ is fixed. The inner integral computes the shared area, the outer integral adds the areas up to compute volume."""
 
-ImageFile(:integral_vector_calculus, imgfile, caption)
+# ImageFile(:integral_vector_calculus, imgfile, caption)
+nothing
 ```
 
+![Figure 14.2 of Strang illustrating the slice when either $x$ is fixed or $y$ is fixed. The inner integral computes the shared area, the outer integral adds the areas up to compute volume.](./figures/strang-slicing.png)
+
 In [Volumes](../integrals/volumes_slice.html) the formula for a volume with a known cross-sectional area is given by $V = \int_a^b CA(x) dx$. The inner integral, $\int_{R_x} f(x,y) dy$ is a function depending on $x$ that yields the area of the slice (where $R_x$ is the region sliced by the line of constant $x$ value). This is consistent with Euler's view of the iterated integral.
 
 
@@ -925,12 +936,12 @@ In [Katz](http://www.jstor.org/stable/2689856) a review of the history of "chang
 We view $R$ in two coordinate systems $(x,y)$ and $(u,v)$. We have that
 
 
-$$
-\begin{align}
+
+\begin{align*}
 dx &= A du + B dv\\
 dy &= C du + D dv,
-\end{align}
-$$
+\end{align*}
+
 
 where $A = \partial{x}/\partial{u}$, $B = \partial{x}/\partial{v}$, $C= \partial{y}/\partial{u}$, and $D = \partial{y}/\partial{v}$. Lagrange, following Euler, first sets $x$ to be constant (as is done in iterated integration). Hence, $dx = 0$ and so $du = -C(B/A) dv$ and, after substitution,  $dy = (D-C(B/A))dv$. Then Lagrange set $y$ to be a constant, so $dy = 0$ and hence $dv=0$ so $dx = Adu$. The area "element" $dx dy = A du \cdot (D - (B/A)) dv = (AD - BC) du dv$. Since areas and volumes are non-negative, the absolute value is used. With this, we have "$dxdy = |AD-BC|du dv$" as the analog of $dx = g'(u) du$.
 
@@ -938,12 +949,11 @@ where $A = \partial{x}/\partial{u}$, $B = \partial{x}/\partial{v}$, $C= \partial
 The expression $AD - BC$ was also derived by Euler, by related means. Lagrange extended the analysis to 3 dimensions. Before doing so, it is helpful to understand the problem from a geometric perspective. Euler was attempting to understand the effects of the following change of variable:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 x &= a + mt + \sqrt{1-m^2} v\\
 y & = b + \sqrt{1-m^2}t -mv
-\end{align}
-$$
+\end{align*}
 
 Euler knew this to be a clockwise *rotation* by an angle $\theta$ with $\cos(\theta) = m$,  a *reflection* through the $x$ axis, and a translation by $\langle a, b\rangle$. All these *should* preserve the area represented by $dx dy$, so he was *expecting* $dx dy = dt dv$.
 
@@ -953,9 +963,12 @@ Euler knew this to be a clockwise *rotation* by an angle $\theta$ with $\cos(\th
 #| echo: false
 imgfile ="figures/euler-rotation.png"
 caption = "Figure from Katz showing rotation of Euler."
-ImageFile(:integral_vector_calculus, imgfile, caption)
+# ImageFile(:integral_vector_calculus, imgfile, caption)
+nothing
 ```
 
+![Figure from Katz showing rotation of Euler.](figures/euler-rotation.png)
+
 The figure, taken from Katz, shows the translation, and rotation that should preserve area on a differential scale.
 
 
@@ -1047,28 +1060,25 @@ The arrows are the images of the standard unit vectors. We see some transformati
 
 
 $$
-\| \det\left(\left[
-\begin{array}{}
+\| \det\left(
+\begin{bmatrix}
 \hat{i} & \hat{j} & \hat{k}\\
 u_1 & u_2 & 0\\
 v_1 & v_2 & 0
-\end{array}
-\right]
+\end{bmatrix}
 \right) \|
 =
-\| \hat{k} \det\left(\left[
-\begin{array}{}
+\| \hat{k} \det\left(
+\begin{bmatrix}
 u_1 & u_2\\
 v_1 & v_2
-\end{array}
-\right]
+\end{bmatrix}
 \right) \|
-= | \det\left(\left[
-\begin{array}{}
+= | \det\left(
+\begin{bmatrix}
 u_1 & u_2\\
 v_1 & v_2
-\end{array}
-\right]
+\end{bmatrix}
 \right)|.
 $$
 
@@ -1108,12 +1118,11 @@ We have [seen](../differentiable_vector_calculus/polar_coordinates.html) how to
 
 
 $$
-J_G = \left[
-\begin{array}{}
+J_G =
+\begin{bmatrix}
 \cos(\theta) & - r\sin(\theta)\\
 \sin(\theta) & r\cos(\theta)
-\end{array}
-\right],
+\end{bmatrix},
 $$
 
 with determinant $r$.
@@ -1156,24 +1165,20 @@ Some [transformations](https://en.wikipedia.org/wiki/Transformation_matrix#Examp
 
 
 $$
-\left[
-\begin{array}{}
+\begin{bmatrix}
 x\\
 y
-\end{array}
-\right] =
-\left[
-\begin{array}{}
+\end{bmatrix}
+=
+\begin{bmatrix}
 a & b\\
 c & d
-\end{array}
-\right]
-\left[
-\begin{array}{}
+\end{bmatrix}
+\begin{bmatrix}
 u\\
 v
-\end{array}
-\right],
+\end{bmatrix}
+,
 $$
 
 or $G(u,v) = \langle au+bv, cu+dv\rangle$. The Jacobian of this *linear* transformation is the matrix itself.
@@ -1219,12 +1224,10 @@ showG(G)
 
 $$
 \frac{1}{\| \vec{l} \|^2}
-\left[
-\begin{array}{}
+\begin{bmatrix}
 l_x^2 - l_y^2 & 2 l_x l_y\\
 2l_x l_y & l_y^2 - l_x^2
-\end{array}
-\right]
+\end{bmatrix}
 $$
 
 For some simple cases: $\langle l_x, l_y \rangle = \langle 1, 1\rangle$, the diagonal, this is $G(u,v) = (1/2) \langle 2v, 2u \rangle$; $\langle l_x, l_y \rangle = \langle 0, 1\rangle$ (the $y$-axis) this is $G(u,v) = \langle -u, v\rangle$.
@@ -1268,12 +1271,10 @@ The determinant of the Jacobian is
 
 $$
 \det(J_G) = \det\left(
-\left[
-\begin{array}{}
+\begin{bmatrix}
 1 & 0\\
 v & u
-\end{array}
-\right]
+\end{bmatrix}
 \right) = u.
 $$
 
@@ -1304,13 +1305,13 @@ What about other triangles, say the triangle bounded by $x=0$, $y=0$ and $y-x=1$
 This can be seen as a reflection through the line $x=1/2$ of the triangle above. If $G_1$ represents the mapping from $U [0,1]\times[0,1]$ into the triangle of the last problem, and $G_2$ represents the reflection through the line $x=1/2$, then the transformation $G_2 \circ G_1$ will map the box $U$ into the desired region. By the chain rule, we have:
 
 
-$$
+
 \begin{align*}
 \int_{(G_2\circ G_1)(U))} f dx &= \int_U (f\circ G_2 \circ G_1) |\det(J_{G_2 \circ G_1}| du \\
 &=
 \int_U (f\circ G_2 \circ G_1) |\det(J_{G_2}(G_1(u))||\det J_{G_1}(u)| du.
 \end{align*}
-$$
+
 
 (In  [Katz](http://www.jstor.org/stable/2689856) it is mentioned that Jacobi showed this in 1841.)
 
@@ -1449,7 +1450,7 @@ $$
 \int_{r=0}^1 r^2 dr \int_{\theta=0}^{\pi}\sin(\theta) = \frac{1}{3} \cdot 2.
 $$
 
-The third equals sign uses separability. The answer for $\bar{ is this value divided by the area, or $2/(3\pi)$.
+The third equals sign uses separability. The answer for $\bar{y}$ is this value divided by the area, or $2/(3\pi)$.
 
 
 ##### Example: Moment of inertia
@@ -1656,9 +1657,12 @@ $$
 imgfile = "figures/spherical-coordinates.png"
 caption = "Figure showing the parameterization by spherical coordinates. (Wikipedia)"
 
-ImageFile(:integral_vector_calculus, imgfile, caption)
+# ImageFile(:integral_vector_calculus, imgfile, caption)
+nothing
 ```
 
+![Figure showing the parameterization by spherical coordinates. (Wikipedia)](./figures/spherical-coordinates.png)
+
 The Jacobian can be computed to be $\rho^2\sin(\phi)$.
 
 

quarto/integral_vector_calculus/line_integrals.qmd

@@ -611,9 +611,13 @@ imgfile = "figures/kapoor-cloud-gate.jpg"
 caption = """
 The Anish Kapoor sculpture Cloud Gate maps the Cartesian grid formed by its concrete resting pad onto a curved surface showing the local distortions.  Knowing the areas of the reflected grid after distortion would allow the computation of the surface area of the sculpture through addition. (Wikipedia)
 """
-ImageFile(:integral_vector_calculus, imgfile, caption)
+#ImageFile(:integral_vector_calculus, imgfile, caption)
+nothing
 ```
 
+![The Anish Kapoor sculpture Cloud Gate maps the Cartesian grid formed by its concrete resting pad onto a curved surface showing the local distortions.  Knowing the areas of the reflected grid after distortion would allow the computation of the surface area of the sculpture through addition. (Wikipedia)
+](./figures/kapoor-cloud-gate.jpg)
+
 We next turn attention to a generalization of line integrals to surface integrals. Surfaces were described in one of three ways: directly through a function as $z=f(x,y)$, as a level curve through $f(x,y,z) = c$, and parameterized through a function $\Phi: R^2 \rightarrow R^3$. The level curve description is locally a function description, and the function description leads to a parameterization ($\Phi(u,v) = \langle u,v,f(u,v)\rangle$) so we restrict to the parameterized case.
 
 

quarto/integral_vector_calculus/review.qmd

@@ -98,13 +98,13 @@ In dimension $m=3$, the **binormal** vector, $\hat{B}$, is the unit vector $\hat
 The [Frenet-Serret]() formulas define the **curvature**, $\kappa$,  and the **torsion**, $\tau$, by
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \frac{d\hat{T}}{ds} &=  & \kappa \hat{N} &\\
 \frac{d\hat{N}}{ds} &= -\kappa\hat{T} & & + \tau\hat{B}\\
 \frac{d\hat{B}}{ds} &= & -\tau\hat{N}&
-\end{align}
-$$
+\end{align*}
+
 
 These formulas apply in dimension $m=2$ with $\hat{B}=\vec{0}$.
 
@@ -121,14 +121,14 @@ The chain rule says $(\vec{r}(g(t))' = \vec{r}'(g(t)) g'(t)$.
 A scalar function, $f:R^n\rightarrow R$, $n > 1$ has a **partial derivative** defined. For $n=2$, these are:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \frac{\partial{f}}{\partial{x}}(x,y) &=
 \lim_{h\rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h}\\
 \frac{\partial{f}}{\partial{y}}(x,y) &=
 \lim_{h\rightarrow 0} \frac{f(x,y+h)-f(x,y)}{h}.
-\end{align}
-$$
+\end{align*}
+
 
 The generalization to $n>2$ is clear - the partial derivative in $x_i$ is the derivative of $f$ when the *other* $x_j$ are held constant.
 
@@ -198,14 +198,13 @@ For $F=\langle f_1, f_2, \dots, f_m\rangle$ the total derivative is the  **Jacob
 
 
 $$
-J_f = \left[
-\begin{align}{}
+J_f =
+\begin{bmatrix}
 \frac{\partial f_1}{\partial x_1} &\quad \frac{\partial f_1}{\partial x_2} &\dots&\quad\frac{\partial f_1}{\partial x_n}\\
 \frac{\partial f_2}{\partial x_1} &\quad \frac{\partial f_2}{\partial x_2} &\dots&\quad\frac{\partial f_2}{\partial x_n}\\
 &&\vdots&\\
 \frac{\partial f_m}{\partial x_1} &\quad \frac{\partial f_m}{\partial x_2} &\dots&\quad\frac{\partial f_m}{\partial x_n}
-\end{align}
-\right].
+\end{bmatrix}.
 $$
 
 This can be viewed as being comprised of row vectors, each being the individual gradients; or as column vectors each being the vector of partial derivatives for a given variable.
@@ -225,7 +224,7 @@ A scalar function $f:R^n \rightarrow R$ and a parameterized curve $\vec{r}:R\rig
 
 
 $$
-d_f(\vec{r}) d_\vec{r} = \nabla{f}(\vec{r}(t))' \vec{r}'(t) =
+d_f(\vec{r}) d\vec{r} = \nabla{f}(\vec{r}(t))' \vec{r}'(t) =
 \nabla{f}(\vec{r}(t)) \cdot \vec{r}'(t),
 $$
 
@@ -356,29 +355,29 @@ $$
 In two dimensions, we have the following interpretations:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \iint_R dA &= \text{area of } R\\
 \iint_R \rho dA &= \text{mass with constant density }\rho\\
 \iint_R \rho(x,y) dA &= \text{mass of region with density }\rho\\
 \frac{1}{\text{area}}\iint_R x \rho(x,y)dA &= \text{centroid of region in } x \text{ direction}\\
 \frac{1}{\text{area}}\iint_R y \rho(x,y)dA &= \text{centroid of region in } y \text{ direction}
-\end{align}
-$$
+\end{align*}
+
 
 In three dimensions, we have the following interpretations:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \iint_VdV &= \text{volume of } V\\
 \iint_V \rho dV &= \text{mass with constant density }\rho\\
 \iint_V \rho(x,y) dV &= \text{mass of volume with density }\rho\\
 \frac{1}{\text{volume}}\iint_V x \rho(x,y)dV &= \text{centroid of volume in } x \text{ direction}\\
 \frac{1}{\text{volume}}\iint_V y \rho(x,y)dV &= \text{centroid of volume in } y \text{ direction}\\
 \frac{1}{\text{volume}}\iint_V z \rho(x,y)dV &= \text{centroid of volume in } z \text{ direction}
-\end{align}
-$$
+\end{align*}
+
 
 To compute integrals over non-box-like regions, Fubini's theorem may be utilized. Alternatively, a **transformation** of variables
 

quarto/integral_vector_calculus/stokes_theorem.qmd

@@ -270,12 +270,12 @@ r(t) = [a*cos(t),b*sin(t)]
 To compute the area of the triangle with vertices $(0,0)$, $(a,0)$ and $(0,b)$ we can orient the boundary counter clockwise. Let $A$ be the line segment from $(0,b)$ to $(0,0)$, $B$ be the line segment from $(0,0)$ to $(a,0)$, and $C$ be the other. Then
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \frac{1}{2} \int_A F\cdot\hat{T} ds &=\frac{1}{2} \int_A -ydx = 0\\
 \frac{1}{2} \int_B F\cdot\hat{T} ds &=\frac{1}{2} \int_B xdy = 0,
-\end{align}
-$$
+\end{align*}
+
 
 as on $A$, $y=0$ and $dy=0$ and on $B$, $x=0$ and $dx=0$.
 
@@ -310,7 +310,7 @@ For the two dimensional case the curl is a scalar. *If* $F = \langle F_x, F_y\ra
 Now assume $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y} = 0$. Let $P$ and $Q$ be two points in the plane. Take any path, $C_1$ from $P$ to $Q$ and any return path, $C_2$, from $Q$ to $P$ that do not cross and such that $C$, the concatenation of the two paths, satisfies Green's theorem. Then, as $F$ is continuous on an open interval containing $D$, we have:
 
 
-$$
+
 \begin{align*}
 0 &= \iint_D 0 dA \\
 &=
@@ -320,7 +320,7 @@ $$
 &=
 \int_{C_1} F \cdot \hat{T} ds + \int_{C_2}F \cdot \hat{T} ds.
 \end{align*}
-$$
+
 
 Reversing $C_2$ to go from $P$ to $Q$, we see the two work integrals are identical, that is the field is conservative.
 
@@ -338,14 +338,14 @@ For example, let $F(x,y) = \langle \sin(xy), \cos(xy) \rangle$. Is this a conser
 We can check by taking partial derivatives. Those of interest are:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \frac{\partial{F_y}}{\partial{x}} &= \frac{\partial{(\cos(xy))}}{\partial{x}} =
 -\sin(xy) y,\\
 \frac{\partial{F_x}}{\partial{y}} &= \frac{\partial{(\sin(xy))}}{\partial{y}} =
 \cos(xy)x.
-\end{align}
-$$
+\end{align*}
+
 
 It is not the case that  $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y}=0$, so this vector field is *not* conservative.
 
@@ -416,25 +416,25 @@ p
 Let $A$ label the red line, $B$ the green curve, $C$ the blue line, and $D$ the black line. Then the area is given from Green's theorem by considering half of the the line integral of $F(x,y) = \langle -y, x\rangle$ or $\oint_C (xdy - ydx)$. To that matter we have:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \int_A (xdy - ydx) &= a f(a)\\
 \int_C (xdy - ydx) &= b(-f(b))\\
 \int_D (xdy - ydx) &= 0\\
-\end{align}
-$$
+\end{align*}
+
 
 Finally the integral over $B$, using integration by parts:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \int_B F(\vec{r}(t))\cdot \frac{d\vec{r}(t)}{dt} dt &=
 \int_b^a \langle -f(t),t)\rangle\cdot\langle 1, f'(t)\rangle dt\\
 &= \int_a^b f(t)dt - \int_a^b tf'(t)dt\\
 &= \int_a^b f(t)dt - \left(tf(t)\mid_a^b - \int_a^b f(t) dt\right).
-\end{align}
-$$
+\end{align*}
+
 
 Combining, we have after cancellation $\oint (xdy - ydx) = 2\int_a^b f(t) dt$, or after dividing by $2$ the signed area under the curve.
 
@@ -469,16 +469,16 @@ The cut leads to a counter-clockwise orientation  on the outer ring and a clockw
 To see that the area integral of $F(x,y) = (1/2)\langle -y, x\rangle$ produces the area for this orientation we have, using $C_1$ as the outer ring, and $C_2$ as the inner ring:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \oint_{C_1} F \cdot \hat{T} ds &=
 \int_0^{2\pi} (1/2)(2)\langle -\sin(t), \cos(t)\rangle \cdot (2)\langle-\sin(t), \cos(t)\rangle dt \\
 &= (1/2) (2\pi) 4 = 4\pi\\
 \oint_{C_2} F \cdot \hat{T} ds &=
 \int_{0}^{2\pi}  (1/2) \langle \sin(t), \cos(t)\rangle \cdot \langle-\sin(t), -\cos(t)\rangle dt\\
 &= -(1/2)(2\pi) = -\pi.
-\end{align}
-$$
+\end{align*}
+
 
 (Using $\vec{r}(t) = 2\langle \cos(t), \sin(t)\rangle$ for the outer ring and $\vec{r}(t) = 1\langle \cos(t), -\sin(t)\rangle$ for the inner ring.)
 
@@ -716,9 +716,13 @@ imgfile ="figures/jiffy-pop.png"
 caption ="""
 The Jiffy Pop popcorn design has a top surface that is designed to expand to accommodate the popped popcorn. Viewed as a surface, the surface area grows, but the boundary - where the surface meets the pan - stays the same. This is an example that many different surfaces can have the same bounding curve. Stokes' theorem will relate a surface integral over the surface to a line integral about the bounding curve.
 """
-ImageFile(:integral_vector_calculus, imgfile, caption)
+# ImageFile(:integral_vector_calculus, imgfile, caption)
+nothing
 ```
 
+![The Jiffy Pop popcorn design has a top surface that is designed to expand to accommodate the popped popcorn. Viewed as a surface, the surface area grows, but the boundary - where the surface meets the pan - stays the same. This is an example that many different surfaces can have the same bounding curve. Stokes' theorem will relate a surface integral over the surface to a line integral about the bounding curve.
+](./figures/jiffy-pop.png)
+
 Were the figure of Jiffy Pop popcorn animated, the surface of foil would slowly expand due to pressure of popping popcorn until the popcorn was ready. However, the boundary would remain the same. Many different surfaces can have the same boundary. Take for instance the upper half unit sphere in $R^3$ it having the curve $x^2 + y^2 = 1$ as a boundary curve. This is the same curve as the surface of the cone $z = 1 - (x^2 + y^2)$ that lies above the $x-y$ plane. This would also be the same curve as the surface formed by a Mickey Mouse glove if the collar were scaled and positioned onto the unit circle.
 
 
@@ -736,7 +740,7 @@ $$
 This gives the series of approximations:
 
 
-$$
+
 \begin{align*}
 \oint_C F\cdot\hat{T} ds &=
 \sum \oint_{C_i} F\cdot\hat{T} ds \\
@@ -747,7 +751,7 @@ $$
 &\approx
 \iint_S \nabla\times{F}\cdot\hat{N} dS.
 \end{align*}
-$$
+
 
 In terms of our expanding popcorn, the boundary integral - after accounting for cancellations, as in Green's theorem - can be seen as a microscopic sum of boundary integrals each of which is approximated by a term $\nabla\times{F}\cdot\hat{N} \Delta{S}$ which is viewed as a Riemann sum approximation for the the integral of the curl over the surface. The cancellation depends on a proper choice of orientation, but with that we have:
 
@@ -1127,13 +1131,13 @@ The divergence theorem provides two means to compute a value, the point here is
 Following Schey, we now consider a continuous analog to the crowd counting problem through a flow with a non-uniform density that may vary in time. Let $\rho(x,y,z;t)$ be the time-varying density and $v(x,y,z;t)$ be a vector field indicating the direction of flow. Consider some three-dimensional volume, $V$, with boundary $S$ (though two-dimensional would also be applicable). Then these integrals have interpretations:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \iiint_V \rho dV &&\quad\text{Amount contained within }V\\
 \frac{\partial}{\partial{t}} \iiint_V \rho dV &=
 \iiint_V \frac{\partial{\rho}}{\partial{t}} dV &\quad\text{Change in time of amount contained within }V
-\end{align}
-$$
+\end{align*}
+
 
 Moving the derivative inside the integral requires an assumption of continuity. Assume the material is *conserved*, meaning that if the amount in the volume $V$ changes it must  flow in and out through the boundary. The flow out through $S$, the boundary of $V$, is