# 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

 $\int_{S}f(u,v)\,d^{2}A=\int f(u,v)\sqrt{c^{2}+u^{2}}\>du\,dv$