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=cv

(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&u\\cos v\\\\\\cos v&-u\\sin v\\end{matrix}\\right|=-u

\\frac{\\partial(y,z)}{\\partial(u,v)}=\\left|\\begin{matrix}\\cos v&-u\\sin v\\\\0&c\\end{matrix}\\right|=c\\cos v

\\frac{\\partial(z,x)}{\\partial(u,v)}=\\left|\\begin{matrix}0&c\\\\\\sin v&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}}\\>du\\,dv

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