From: W. Trevor King <wking@drexel.edu>
Date: Wed, 4 Apr 2012 19:25:14 +0000 (-0400)
Subject: Fix dA -> \dd A in Serway and Jewett v8's problem 24.13.
X-Git-Url: http://git.tremily.us/?p=course.git;a=commitdiff_plain;h=79b90a86def33abb32a6a3de7ef9aa58b8dac339

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 5a83d2b..98034e8 100644
--- 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,