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<record version="4" id="6666">
 <title>example of integration with respect to surface area on a helicoid</title>
 <name>ExampleOfIntegrationWithRespectToSurfaceArea</name>
 <created>2005-01-27 00:19:44</created>
 <modified>2005-05-04 04:26:58</modified>
 <type>Example</type>
<parent id="6660">surface integration with respect to area</parent>
 <creator id="6075" name="rspuzio"/>
 <author id="6075" name="rspuzio"/>
 <classification>
	<category scheme="msc" code="28A75"/>
 </classification>
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 <content>In this example, we shall consider itegration with respect to surface area on the helicoid.

The helicoid may be parameterized as follows:
 $$x = u \sin v$$
 $$y = u \cos v$$
 $$z = c v$$
(The constant $c$ may be thought of as the ``pitch of the screw''.)  Computing derivatives and appying trigonometric identities, we obtain
$$\frac{\partial (x, y)}{\partial (u,v)} =
\left| \begin{matrix}
\sin v &amp; u \cos v \\
\cos v &amp; - u \sin v
\end{matrix} \right| =
- u$$
$$\frac{\partial (y, z)}{\partial (u,v)} =
\left| \begin{matrix}
\cos v &amp; - u \sin v \\
0 &amp; c
\end{matrix} \right| =
c \cos v$$
$$\frac{\partial (z, x)}{\partial (u,v)} =
\left| \begin{matrix}
0 &amp; c \\
\sin v &amp; u \cos v
\end{matrix} \right| =
- c \sin v.$$
From this we have
 $$\sqrt{ \left(  \frac{\partial (x,y)}{\partial (u,v)} \right)^2 +  \left( \frac{\partial (y,z)}{\partial (u,v)} \right)^2 + \left( \frac{\partial (z,x)}{\partial (u,v)} \right)^2 } =$$
 $$\sqrt{ u^2 + c^2 \cos^2 v + c^2 \sin^2 v } = \sqrt { u^2 + c^2 }$$
so we can compute area integrals over helicoids as follows
 $$\int_S f(u,v) \, d^2 A = \int f(u,v) \sqrt{ c^2 + u^2 } \&gt; du \, dv$$

To return to the main entry \PMlinkid{click here}{6660}</content>
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