Find the parametric equation of the surface cut from a sphere $x^2+y^2+z^2=16$ by a cone $z=\sqrt{x^2+y^2}$.
Attempt: I want to use spherical coordinates. Solving the sphere and cone, the surface is given by $$S:z=\sqrt{16-x^2-y^2}, ~~ x^2+y^2\leq 8.$$
A parametrization is given by
$$r(u,v)=(4 \,cos \,u\,sin\, v, 4 \,sin \,u\,sin \,v, 4 \,cos \,v)$$ where $u\in[0,2\pi]$.
Question: If my argument is correct, what is the range of $v$ and how to find it. If possible, please provide me a figure by marking the surface cut.
Is the range of $v$ is $[0,\pi/4]$, since, $4\cos\, v=z=\sqrt{x^2+y^2}=\sqrt{16\sin^2\, v\cos^2\, u+16\sin^2\, v\sin^2\, u}=4\sin\, v$ implying $tan\, v=1$.
