align fix; theorem style; condition number

quarto/integrals/surface_area.qmd

@@ -50,19 +50,24 @@ 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
->
-> $$
-> \int_a^b 2\pi f(x) \cdot \sqrt{1 + f'(x)^2} dx.
-> $$
->
-> If the curve is parameterized by $(g(t), f(t))$ between $a$ and $b$ then the surface area is
->
-> $$
-> \int_a^b 2\pi f(t) \cdot \sqrt{g'(t)^2 + f'(t)^2} dx.
-> $$
->
-> These formulas do not add in the surface area of either of the ends.
+
+::: {.callout-note icon=false}
+## Surface area of a rotated curve
+
+The surface area generated by rotating the graph of $f(x)$ between $a$ and $b$  about the $x$-axis is given by the integral
+
+$$
+\int_a^b 2\pi f(x) \cdot \sqrt{1 + f'(x)^2} dx.
+$$
+
+If the curve is parameterized by $(g(t), f(t))$ between $a$ and $b$ then the surface area is
+
+$$
+\int_a^b 2\pi f(t) \cdot \sqrt{g'(t)^2 + f'(t)^2} dx.
+$$
+
+These formulas do not add in the surface area of either of the ends.
+:::
 
 
 
@@ -129,7 +134,7 @@ Lets see that the surface area of an open cone follows from this formula, even t
 A cone can 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:
 
 
-
+$$
 \begin{align*}
 \int_0^h 2\pi f(x) \sqrt{1 + f'(x)^2}dx
 &= \int_0^h 2\pi x \tan(\theta) \sqrt{1 + \tan(\theta)^2}dx   \\
@@ -137,6 +142,7 @@ A cone can be envisioned as rotating the function $f(x) = x\tan(\theta)$ between
 &= \pi \tan(\theta) \sec(\theta) h^2                          \\
 &= \pi r^2 / \sin(\theta).
 \end{align*}
+$$
 
 
 (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.)
@@ -322,13 +328,14 @@ 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
@@ -378,11 +385,12 @@ 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:
 
 
-
+$$
 \begin{align*}
 \text{cm}_x &= \frac{1}{L} \int_a^b g(t) \sqrt{g'(t)^2 + f'(t)^2} dt\\
 \text{cm}_y &= \frac{1}{L} \int_a^b f(t) \sqrt{g'(t)^2 + f'(t)^2} dt.
 \end{align*}
+$$
 
 
 Here, $L$ is the arc length of the curve.