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Given the parameterization of the unit sphere $x^2+y^2+z^2=1$ as

$x = \displaystyle\frac{u}{\sqrt{1+u^2+v^2}} $

$y = \displaystyle\frac{v}{\sqrt{1+u^2+v^2}} $

$z = \displaystyle\frac{1}{\sqrt{1+u^2+v^2}} $

Find $ds^2=dx^2+dy^2+dz^2$ and using the metric computer the area of the hemisphere $z\geq0$

I got:

$dx = \displaystyle\frac{v^2+1}{(u^2+v^2+1)^{3/2}}du-\displaystyle\frac{uv}{(u^2+v^2+1)^{3/2}}$dv

$dy = -\displaystyle\frac{uv}{(u^2+v^2+1)^{3/2}}du+\displaystyle\frac{u^2+1}{(u^2+v^2+1)^{3/2}}dv$

$dz = \displaystyle\frac{-u}{(u^2+v^2+1)^{3/2}}du-\displaystyle\frac{v}{(u^2+v^2+1)^{3/2}}dv$

And

$dx^2+dy^2+dz^2= \displaystyle\frac{(v^4+2v^2+1+u^2+u^2v^2)du^2+(u^4+2u^2+1+v^2+u^2v^2)dv^2+(-2uv-2u^3v-2uv^3)dudv}{(u^2+v^2+1)^3}$

But I can't see an obvious simplification

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    yes i have to deal with a certain parameterization, but using Sashsa' factoring has given me what i needed to finish the problem, just checking over a few more things though2012-10-02

1 Answers 1

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Factor your coefficients: $ v^4 + 2 v^2 + 1 + u^2 + u^2 v^2 = (1+v^2)(1+u^2+v^2) $ $ u^4 + 2 u^2 + 1 + v^2 + u^2 v^2 = (1+u^2) (1+u^2+v^2) $ $ (-2 u v-2 u^2 v-2 u v^3) = - 2u v(1+u^2+v^2) $ Then cancel common factors of the numerator and the denominator.

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    If you mean to integrate over the hemisphere, then $u$ and $v$ are unconstrained, that integrate over $(u,v) \in \mathbb{R}^2$.2012-10-02