surface area of a sphere above a cylinder I need to find the surface area of the sphere $x^2+y^2+z^2=4$ above the cone $z = \sqrt{x^2+y^2}$, but I'm not sure how. I know that the surface area of a surface can be calculated with the equation $A=\int{\int_D{\sqrt{f_x^2+f_y^2+1}}}dA$, but I'm not sure how to take into account the constraint that it must lie above the cone. How is this done?

If you think about it in the $xz$ plane, the sphere becomes a circle of radius $2$ and the cone becomes the pair of lines $z=x$ and $z=-x$.  These intersect at $x=z=\sqrt 2$ and at $x=-\sqrt 2, z=\sqrt 2$.  So you are looking for the part of the sphere above $z=\sqrt 2$.  In spherical coordinates, that is $0 \le \theta \le \frac \pi 4$
