integration with respect to surface area on a helicoid

To illustrate the result derived in http://planetmath.org/node/6666example 3, 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 .) The integral we need to compute in this case is

$$A=\int d^{2}A={\int }_0^{h/c}{\int }_0^r\sqrt{c^2+u^2}𝑑u𝑑v=$$

$${\int }_0^{h/c}\left(\frac{1}{2}r\sqrt{c^2+r^2}+\frac{c^2}{2}\log \left\{\frac{r}{c}+\sqrt{1+{\left(\frac{r}{c}\right)}^2}\right\}\right)𝑑v=$$

$$\frac{rh}{2}\sqrt{1+{\left(\frac{r}{c}\right)}^2}+\frac{ch}{2}\log \left\{\frac{r}{c}+\sqrt{1+{\left(\frac{r}{c}\right)}^2}\right\}$$

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:

${\displaystyle \int (x^2+y^2)d^{2}A}=$ ${\displaystyle {\int }_0^{h/c}}{\displaystyle {\int }_0^r}{u}^2\sqrt{c^2+u^2}𝑑u𝑑v=$

${\displaystyle {\int }_0^{h/c}}\left({\displaystyle \frac{r(2r^2+c^2)}{8}}\sqrt{c^2+r^2}-{\displaystyle \frac{c^4}{8}}\log \left\{{\displaystyle \frac{r}{c}}+\sqrt{1+{\left({\displaystyle \frac{r}{c}}\right)}^2}\right\}\right)𝑑v=$

${\displaystyle \frac{rh(2r^2+c^2)}{8}}\sqrt{1+{\left({\displaystyle \frac{r}{c}}\right)}^2}-{\displaystyle \frac{h{c}^3}{8}}\log \left\{{\displaystyle \frac{r}{c}}+\sqrt{1+{\left({\displaystyle \frac{r}{c}}\right)}^2}\right\}$

http://planetmath.org/node/6660main entry http://planetmath.org/node/6666previous example http://planetmath.org/node/6668next example

Title	integration with respect to surface area on a helicoid
Canonical name	IntegrationWithRespectToSurfaceAreaOnAHelicoid
Date of creation	2013-03-22 14:58:04
Last modified on	2013-03-22 14:58:04
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	14
Author	rspuzio (6075)
Entry type	Example
Classification	msc 28A75