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16.tex

@@ -20,6 +20,18 @@ The following graph demonstrates the general idea.
 
 \medskip
 \noindent
-Now imagine that the graph is rotated about the $x$-axis.
+Now imagine that the graph is rotated about the $x$ axis.
 What is the volume subtended by the area under the curve?
 
+\medskip
+\noindent
+To solve, use cylindrical coordinates.
+The $x$ axis becomes the $z$ axis.
+We want $x$ to become a function of $y$ so invert the function.
+$$y={1\over\sqrt{1+x^2}}
+\quad\rightarrow\quad
+{1\over y^2}=1+x^2
+\quad\rightarrow\quad
+x=\sqrt{{1\over y^2}-1}
+$$
+