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Area Element of Elliptic Cylinder Coordinates

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Compute the area element for elliptic cylinder coordinates


    2. Relevant equations
    The coordinates are defined as follows:
    x=a*cosh(u)*cos(v)
    y=a*sinh(u)*sin(v)


    3. The attempt at a solution
    Starting from the assumption that the area element dA=dx*dy, I found dx and dy:
    dx=a*du*sinh(u)*cos(v) - a*cosh(u)*dv*sin(v)
    dy=a*du*cosh(u)*sin(v) - a*sinh(u)*dv*cos(v)

    Then multiplying these together, to get dA:
    dA=[(a^2)*sinh(u)*cosh(u)*sin(v)*cos(v)*(du^2 - dv^2)] +
    [(a^2)*du*dv*((sinh(u))^2)*((cos(v))^2) - (cosh(u))^2)*((sin(v))^2)]

    I don't like this answer for a couple of reasons. It seems like there should be a tidier, more compact expression than what I have. Compared to surface elements in other coordinate systems, this is frankly a mess. Also, I don't think I've seen a "du^2" in any area element formulae either, which I'm not sure makes it wrong, but I feel a little uneasy about it anyway.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 12, 2009 #2

    HallsofIvy

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    Science Advisor

    When you say "multiplying these together", what kind of product did you use? If a surface is given by parametric equations, x= x(u,v), y= y(u,v), z= z(u,v), then the "position vector" is [itex]\vec{r}= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}[/itex], the derivatives in the directions of the u, v axes are [itex]\vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k}[/itex] and [itex]\vec{r}_v= x_v \vec{i}+ y_v\vec{j}+ z_v\vec{k}[/itex]. The differential of surface area is the length of the cross product of those two vectors times dudv.
     
  4. Nov 12, 2009 #3
    So in vector form: dr/du=(a*sinh(u)*cos(v) , a*cosh(u)*sin(v) , 0)
    and dr/dv=(-a*cosh(u)*sin(v) , a*sinh(u)*cos(v) , 0)

    Multiplying across by du and dv respectively:
    dr=(a*sinh(u)*cos(v)du , a*cosh(u)*sin(v)du , 0) with dr w.r.t "u"
    and dr=(-a*cosh(u)*sin(v)dv , a*sinh(u)*cos(v)dv , 0) with dr w.r.t "v"

    Then cross product of these vectors gives:
    dA=length of the vector: [0,0,(a^2*du*dv*(sinh(u)^2*cos(v)^2 + cosh(u)^2*sin(v)^2))]

    And the length of this vector is just the third component, so is this the answer for dA?
    It seems a better answer than my original one, and the reasoning behind this method makes perfect sense to me.

    dA= a^2*du*dv*(sinh(u)^2*cos(v)^2 + cosh(u)^2*sin(v)^2)
     
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