example of integration with respect to surface area on a helicoid


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

∂⁡(x,y)∂⁡(u,v)=|sin⁡vu⁢cos⁡vcos⁡v-u⁢sin⁡v|=-u
∂⁡(y,z)∂⁡(u,v)=|cos⁡v-u⁢sin⁡v0c|=c⁢cos⁡v
∂⁡(z,x)∂⁡(u,v)=|0csin⁡vu⁢cos⁡v|=-c⁢sin⁡v.

From this we have

(∂⁡(x,y)∂⁡(u,v))2+(∂⁡(y,z)∂⁡(u,v))2+(∂⁡(z,x)∂⁡(u,v))2=
u2+c2⁢cos2⁡v+c2⁢sin2⁡v=u2+c2

so we can compute area integrals over helicoids as follows

∫Sf⁢(u,v)⁢d2⁢A=∫f⁢(u,v)⁢c2+u2⁢𝑑u⁢𝑑v

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Title example of integration with respect to surface area on a helicoid
Canonical name ExampleOfIntegrationWithRespectToSurfaceAreaOnAHelicoid
Date of creation 2013-03-22 14:58:01
Last modified on 2013-03-22 14:58:01
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 7
Author rspuzio (6075)
Entry type Example
Classification msc 28A75