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