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!
 A: The cone $z > \sqrt{x^2 + y^2}$ is a cone of a certain angle
$\phi_0$ around the $z$-axis.
It contains a "cap" of the sphere, that is, a portion of the sphere
bounded by a circle.
You can consider this "cap" to consist of all points on the sphere
at an angle up to $\phi_0$ from the $z$-axis (in spherical coordinates),
or you can consider the "cap" to consist of all points
"above" the plane $z=z_0$ for a certain value $z_0.$
Part of the exercise is to figure out what $\phi_0$ or $z_0$ might be.
For $\phi_0,$ it may help to write down how you would compute the
angle $\phi$ (the angle to the $z$-axis) for the spherical coordinates
of a point, given the Cartesian coordinates $(x,y,z)$ of that point.
If you set up your area integral using spherical coordinates,
then you don't need to know $z_0,$ only $\phi_0.$
On the other hand, you can set up an integral in cylindrical or Cartesian
coordinates using $z_0$ and ignoring $\phi_0.$
Personally, I would probably use spherical coordinates for this problem.
A: 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?
