typos

quarto/integrals/surface_area.qmd

@@ -85,7 +85,7 @@ To see why this formula is as it is, we look at the parameterized case, the firs
 Let a partition of $[a,b]$ be given by $a = t_0 < t_1 < t_2 < \cdots < t_n =b$. This breaks the curve into a collection of line segments. Consider the line segment connecting $(g(t_{i-1}), f(t_{i-1}))$ to $(g(t_i), f(t_i))$. Rotating this around the $x$ axis will generate something approximating a disc, but in reality will be the frustum of a cone. What will be the surface area?
 
 
-Consider a right-circular cone parameterized by an angle $\theta$ and the largest radius $r$ (so that the height satisfies $r/h=\tan(\theta)$). If this cone were made of paper, cut up a side, and layed out flat, it would form a sector of a circle, whose area would be $R^2\gamma/2$ where $R$ is the radius of the circle (also the side length of our cone), and $\gamma$ an angle that we can figure out from $r$ and $\theta$. To do this, we note that the arc length of the circle's edge is $R\gamma$ and also the circumference of the bottom of the cone so $R\gamma = 2\pi r$. With all this, we can solve to get $A = \pi r^2/\sin(\theta)$. But we have a frustum of a cone with radii $r_0$ and $r_1$, so the surface area is a difference: $A = \pi (r_1^2 - r_0^2) /\sin(\theta)$.
+Consider a right-circular cone parameterized by an angle $\theta$ and the largest radius $r$ (so that the height satisfies $r/h=\tan(\theta)$). If this cone were made of paper, cut up a side, and laid out flat, it would form a sector of a circle, whose area would be $R^2\gamma/2$ where $R$ is the radius of the circle (also the side length of our cone), and $\gamma$ an angle that we can figure out from $r$ and $\theta$. To do this, we note that the arc length of the circle's edge is $R\gamma$ and also the circumference of the bottom of the cone so $R\gamma = 2\pi r$. With all this, we can solve to get $A = \pi r^2/\sin(\theta)$. But we have a frustum of a cone with radii $r_0$ and $r_1$, so the surface area is a difference: $A = \pi (r_1^2 - r_0^2) /\sin(\theta)$.
 
 
 Relating this to our values in terms of $f$ and $g$, we have $r_1=f(t_i)$, $r_0 = f(t_{i-1})$, and $\sin(\theta) = \Delta f / \sqrt{(\Delta g)^2 + (\Delta f)^2}$, where $\Delta f = f(t_i) - f(t_{i-1})$ and similarly for $\Delta g$.