Surface Area of Hellicoid

purplerain · #1 $Old$ November 20th, 2009, 07:44 PM

Find the area of the surface.
The helicoid (or spiral ramp) with vector equation

$r(u,v) = ucos(v) i + usin(v) j + v k\ 0\leq u\leq 1, 0\leq v\leq \pi$

$r_u = cos(v)i+sin(v)j +0k$
$r_v = -cos(v)i + ucos(v)j + k$

Next i found $|r_u \times r_v|$

and got $1 + 2u + u^2$

Im not even sure if this is right, i got confused by the k term (algebra is rusty). Last, i need to take the

$\int\int |r_u \times r_v| dudv$

Im not sure how to take the integral of this, help would be much appreciated.

shawsend · #2 $Old$ November 21st, 2009, 03:13 AM

If $r(u,v)=\{u\cos(v), u\sin(v),v\}$ then $r_u=\{\cos(v),sin(v),0\}$ and $r_v=\{-u\sin(v),u\cos(v),1\}$

Then:

$r_u \times r_v=\left|\begin{array}{ccc}i & j & k \\ \cos(v) & \sin(v) & 0 \\ -u\sin(v) & u\cos(v) & 1 \end{array}\right|=\{\sin(v),-\cos(v),u\}$

so that:

$\mathop\int\int\limits_{\hspace{-15pt} R} |r_u \times r_v| dA=\mathop\int\int\limits_{\hspace{-15pt}R}\sqrt{\sin^2(v)+\cos^2(v)+u^2}\; du dv$