Area of a hyperboloid

  • Thread starter sarah7
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Hi,

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!

thanks
 

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  • #2
jfizzix
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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.
 
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thanks for your reply but how can I get rid of a,b and c
 
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jfizzix
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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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