Prove its Gauss curvature K = 1

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SUMMARY

The discussion centers on proving that the Gauss curvature K equals 1 for a surface defined by the first fundamental form with E = G = 4(1+u²+v²) - 2 and F = 0. The formula used for K is K = \frac{-1}{2\sqrt{EG}}[(\frac{E_v}{\sqrt{EG}})_v + (\frac{G_u}{\sqrt{EG}})_u]. The user initially struggled with the calculations but later identified an error in using the denominators of E and G. The conclusion reached is that careful substitution and differentiation are crucial for deriving the correct curvature value.

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Homework Statement



Assume that the surface has the first fundamental form as

E = G = 4(1+u2+v2)-2

F = 0[/B]

Homework Equations



K = [itex]\frac{-1}{2\sqrt{EG}}[(\frac{E_v}{\sqrt{EG}})_v + (\frac{G_u}{\sqrt{EG}})_u][/itex][/B]

The Attempt at a Solution



Ev = -16v*(1+u2+v2)-3

Gu = -16u*(1+u2+v2)-3


When I take the partials and find K, I get something messy that doesn't lead me to the conclusion that K = 1.[/B]
 
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Start by substituting E=G and \sqrt(EG) = |E|.
What did you end up with?
 
Thanks for the reply. I found my error a little bit ago. For some reason, I kept using the denominator of E and G AS E and G. -_- It was late, so I'll credit it to sleep deprivation. Heh.
 

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