integration with respect to surface area on a helicoid
Example4OfIntegrationWithRespectToSurfaceArea
2005-01-27 00:56:50
2007-05-19 22:12:29
Example
surface integration with respect to area
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To illustrate the result derived in \PMlinkid{example 3}{6666}, let us compute the area of a portion of helicoid of height $h$ and radius $r$. (This calculation will tell us how much material is needed to make an \PMlinkescapetext{Archimedean screw}.) The integral we need to compute in this case is
$$A = \int d^2 A = \int_0^{h/c} \int_0^r \sqrt{ c^2 + u^2 } \> du \, dv = $$
$$\int_0^{h/c} \left( {1 \over 2} r \sqrt{ c^2 + r^2} + {c^2 \over 2} \log \left\{ \frac{r}{c} + \sqrt{ 1 + \left( \frac{r}{c} \right)^2 } \right\} \right) \> dv =$$
$$\frac{rh}{2} \sqrt{ 1 + \left( \frac{r}{c} \right)^2 } + {ch \over 2} \log \left\{ \frac{r}{c} + \sqrt{ 1 + \left( \frac{r}{c} \right)^2 } \right\}$$
As a second illustration, let us compute the second moment of a helicoid about the axis of rotation. In mechanics, this would be called the moment of inertia of the helicoid and determines how much energy is needed to make the screw rotate. This is determined as follows:
\begin{eqnarray*}
\int (x^2 + y^2) d^2 A = &\int_0^{h/c} \int_0^r u^2 \sqrt{ c^2 + u^2 } \> du \, dv = \\
&\int_0^{h/c} \left( \frac{r (2 r^2 + c^2)}{8} \sqrt{ c^2 + r^2 } - \frac{c^4}{8} \log \left\{ \frac{r}{c} + \sqrt{ 1 + \left( \frac{r}{c} \right)^2 } \right\} \right) \, dv = \\
& \frac{r h (2 r^2 + c^2)}{8} \sqrt{ 1 + \left( \frac{r}{c} \right)^2 } - \frac{h c^3}{8} \log \left\{ \frac{r}{c} + \sqrt{ 1 + \left( \frac{r}{c} \right)^2 } \right\}
\end{eqnarray*}
\PMlinkescapetext{\sl Quick links:}
\PMlinkid{main entry}{6660}
\PMlinkid{previous example}{6666}
\PMlinkid{next example}{6668}