Area of the portion of the sphere Find the area of the portion of the sphere of radius 1 (centered at the origin) that is in the cone $$z > \sqrt{x^2 + y^2}.$$
I tried to find the formula by the integral but I still did not get it!

For a sphere, we have $x^2+y^2+z^2=1 \Leftrightarrow z=\sqrt{1-x^2-y^2}$
$f_y=-2y(\frac{1}{2})(1-x^2-y^2)^{-1/2}$, $f_x=-2x(\frac{1}{2})(1-x^2-y^2)^{-1/2}$.
So by the formula for a surface area, we have 
$$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{(-2y(\frac{1}{2})(1-x^2-y^2)^{-1/2})^2+(-2x(\frac{1}{2})(1-x^2-y^2)^{-1/2})^2+1}dydx.$$ 
$$=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left(\frac{x^2+y^2}{1-x^2-y^2}+1\right)^{\frac{1}{2}}dydx.$$
Can you take it from here?
