16.8 + fix some formatting

site/posts/multivariable-thomas-16.1.html

@@ -40,7 +40,7 @@
 </p>
 <p>
     Another example is to find the mass of a circular arc in the first quadrant, with radius 2 and density `x^2 + y^2`. Remember that the parameterization of
-    a circle is `vec c(t) = [rcos(t), rsin(t)]`, and the radial span of the first quadrant is from `0` to `frac pi 2`. Thus, `vec r(t) = [2cos(t), 2sin(t)]` for `0 <= t <= frac pi 2`.
+    a circle is `vec c(t) = \[rcos(t), rsin(t)]`, and the radial span of the first quadrant is from `0` to `frac pi 2`. Thus, `vec r(t) = \[2cos(t), 2sin(t)]` for `0 <= t <= frac pi 2`.
     The density function here is convenient because it's just `r^2`, based on the pythagorean identity. If you substitute in `2cos(t)` for `x` and `2sin(t)` for y, you will get
     `2^2 = 4`. Constant density is always nice! Finally, we need to find the element of integration. `vec r'(t) = \[-2sin(t), 2cos(t)]`, so `|r'|dt` = `sqrt(4sin^2(t) + 4cos^2(t))` -
     `2 dt`. Thus, we have the integral `int_0^{frac pi 2} 8 dt`; this obviously works out to `4pi`.