# Central Projection of the Sphere

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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Spherical coordinates are for spheres. I gather you may not have such freedom here. That said: notice that you can factor the $dudv$ term $-2uv(1+u^2+v^2)$. I'm not totally sure your multiplication is correct (I've not checked it all) – James S. Cook Oct 2 '12 at 1:51
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 though – rckrd Oct 2 '12 at 1:56

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.
After I've calculated $ds^2$ and $da$, the area element, I'm having trouble finding the bounds in which to integrate. – rckrd Oct 2 '12 at 2:05
If you mean to integrate over the hemisphere, then $u$ and $v$ are unconstrained, that integrate over $(u,v) \in \mathbb{R}^2$. – Sasha Oct 2 '12 at 2:28