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Thread: Surface Area of Hellicoid

  1. #1
    Junior Member
    Joined
    Mar 2009
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    63

    Surface Area of Hellicoid

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

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

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

    Next i found $\displaystyle |r_u \times r_v| $

    and got $\displaystyle 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

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

    Im not sure how to take the integral of this, help would be much appreciated.
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  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    If $\displaystyle r(u,v)=\{u\cos(v), u\sin(v),v\}$ then $\displaystyle r_u=\{\cos(v),sin(v),0\}$ and $\displaystyle r_v=\{-u\sin(v),u\cos(v),1\}$

    Then:

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

    so that:

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