Fix dA -> \dd A in Serway and Jewett v8's problem 24.13.

diff --git a/latex/problems/Serway_and_Jewett_8/problem24.13.tex b/latex/problems/Serway_and_Jewett_8/problem24.13.tex
index 5a83d2ba76752392bfcd0784cb92c0c7a8339b72..98034e84eb34e49afaa3d507ba2a347c55686008 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem24.13.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem24.13.tex
@@ -12,7 +12,7 @@ Because the electric field is directed downward, there is no flux
 through the walls of the cylinder.  All the flux crosses the cylinder
 at the end caps.  If the area of the end cap is $A$, that flux is
 \begin{equation}
-  \Phi_E \equiv \oint_S \vect{E}\cdot\vect{dA} = E_{500}A - E_{600}A \;,
+  \Phi_E \equiv \oint_S \vect{E}\cdot\vect{\dd A} = E_{500}A - E_{600}A \;,
 \end{equation}
 where $E_{500}=120\U{N/C}$ and $E_{600}=100\U{N/C}$.  From Gauss's
 law,