make pdf file generation work

quarto/integrals/surface_area.qmd

@@ -34,9 +34,19 @@ revolution, there is an easier way. (Photo credit to
 [firepanjewellery](http://firepanjewellery.com/).)
 """
 
-ImageFile(:integrals, imgfile, caption)
+# ImageFile(:integrals, imgfile, caption)
+nothing
 ```
 
+![The exterior of the Jimi Hendrix Museum in Seattle has the signature
+style of its architect Frank Gehry. The surface is comprised of
+patches. A general method to find the amount of material to cover the
+surface - the surface area - might be to add up the area of *each* of the
+patches. However, in this section we will see for surfaces of
+revolution, there is an easier way. (Photo credit to
+[firepanjewellery](http://firepanjewellery.com/).)
+](./figures/gehry-hendrix.jpg)
+
 > The surface area generated by rotating the graph of $f(x)$ between $a$ and $b$  about the $x$-axis is given by the integral
 >
 > $$
@@ -110,7 +120,6 @@ If we assume integrability of the integrand, then as our partition size goes to
 ```{julia}
 #| hold: true
 #| echo: false
-#| cache: true
 ## {{{approximate_surface_area}}}
 
 xs,ys = range(-1, stop=1, length=50), range(-1, stop=1, length=50)
@@ -154,7 +163,7 @@ Lets see that the surface area of an open cone follows from this formula, even t
 A cone be be envisioned as rotating the function $f(x) = x\tan(\theta)$ between $0$ and $h$ around the $x$ axis. This integral yields the surface area:
 
 
-$$
+
 \begin{align*}
 \int_0^h 2\pi f(x) \sqrt{1 + f'(x)^2}dx
 &= \int_0^h 2\pi x \tan(\theta) \sqrt{1 + \tan(\theta)^2}dx   \\
@@ -162,7 +171,7 @@ $$
 &= \pi \tan(\theta) \sec(\theta) h^2                          \\
 &= \pi r^2 / \sin(\theta).
 \end{align*}
-$$
+
 
 (There are many ways to express this, we used $r$ and $\theta$ to match the work above. If the cone is parameterized by a height $h$ and radius $r$, then the surface area of the sides is $\pi r\sqrt{h^2 + r^2}$. If the base is included, there is an additional $\pi r^2$ term.)
 
@@ -350,14 +359,14 @@ plot(g, f, 0, 1pi)
 The integrand simplifies to $8\sqrt{2}\pi \sin(t) (1 + \cos(t))^{3/2}$. This lends itself to $u$-substitution with $u=\cos(t)$.
 
 
-$$
+
 \begin{align*}
 \int_0^\pi 8\sqrt{2}\pi \sin(t) (1 + \cos(t))^{3/2}
 &= 8\sqrt{2}\pi \int_1^{-1} (1 + u)^{3/2} (-1) du\\
 &= 8\sqrt{2}\pi (2/5) (1+u)^{5/2} \big|_{-1}^1\\
 &= 8\sqrt{2}\pi (2/5) 2^{5/2} = \frac{2^7 \pi}{5}.
 \end{align*}
-$$
+
 
 ## The first Theorem of Pappus
 
@@ -406,12 +415,12 @@ surface(ws..., legend=false, zlims=(-12,12))
 The surface area of sphere will be SA$=2\pi \rho (\pi r) = 2 \pi^2 r \cdot \rho$. What is $\rho$? The centroid of an arc formula can be derived in a manner similar to that of the centroid of a region. The formulas are:
 
 
-$$
-\begin{align}
+
+\begin{align*}
 \text{cm}_x &= \frac{1}{L} \int_a^b g(t) \sqrt{g'(t)^2 + f'(t)^2} dt\\
 \text{cm}_y &= \frac{1}{L} \int_a^b f(t) \sqrt{g'(t)^2 + f'(t)^2} dt.
-\end{align}
-$$
+\end{align*}
+
 
 Here, $L$ is the arc length of the curve.