surface area of a sphere above a cylinder

Question

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?

Answer 1

This problem is setup for spherical coordinates, so I would recommend you don't use the surface area formula you've written in terms of Cartesian coordinates.

Since $r=2$ is fixed, the infinitesimal area element in spherical coordinates is $dA=4\sin\theta d\theta d\phi$. The spherical cap is bounded by $0<\theta<\pi/4$ and $0<\phi<2\pi$. So try to evaluate the integral:

$$\int_0^{2\pi}\int_{0}^{\frac{\pi}{4}}4\sin\theta d\theta d\phi$$

Answer 2

Hint: Use spherical coordinates. $dA = r^2\sin\theta d\theta d\phi$ with $0<\theta<\pi$. The surface area becomes $\iint_D dA$.

Answer 3

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 $0 \le \theta \le \frac \pi 4$