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

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.
