integration with respect to surface area on a helicoid

To illustrate the result derived in 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


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:

(x2+y2)d2A= 0h/c0ru2c2+u2𝑑u𝑑v=
rh(2r2+c2)81+(rc)2-hc38log{rc+1+(rc)2} entry example 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