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Area of a hyperboloid

  1. Oct 5, 2013 #1

    I wanted to ask about how to compute the area of a hyperboloid given in general by

    (x/a)^2 + (y/b)^2 - (z/c)^2 = 1

    I know this is parameterised by (acos(u)cosh(v), bsin(u)cosh(v), csinh(v))
    and I used the definition that A=∫√(EG-F^2) where E,G,F are from the first fundamental form
    however, I wasn't able to integrate this as it is very complicated.
    I thought there might be a way of solving this by considering the area of a tiny parallelogram and then integrating it but I wasn't sure how to start that!

  2. jcsd
  3. Oct 5, 2013 #2


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    I would perform a coordinate transformation to get rid of a,b, and c. The resulting hyperboloid will be a surface of revolution, simplifying the integral to be over one variable.
  4. Oct 6, 2013 #3
    thanks for your reply but how can I get rid of a,b and c
  5. Oct 6, 2013 #4


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    x' = x/a
    y' = y/b
    z' = z/c

    Then x'^2 + y'^2-z'^2 =1
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