The part of the sphere x2 + y2 + z2 = 16 that lies above the cone z = x2 + y2 . (Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of u and/or v.) where z > x2 + y2?

Answer:(x,y,z)=(ucos(v), usin(v), [tex]\sqrt{16-u^{2}[/tex])where [tex]0\leq u\leq 2\sqrt{2}[/tex] and [tex]0\leq v\leq 2\pi[/tex]Step-by-step explanation:Equation of a cone is [tex]z=\sqrt{x^{2} +y^{2}}[/tex]Equation of a paraboloid is [tex]z=x^{2} +y^{2}[/tex]I have parametrised cone here. Please note that equation for cone in the question, is actually a paraboloid.Imagine a sphere of radius 4, centered at origin and intersecting a cone also centered at origin and height along positive z-axis, given by the equations[tex]x^{2} +y^{2} +z^{2} = 16\\z=\sqrt{x^{2} +y^{2}}[/tex]where [tex]z\geq x^{2} +y^{2}[/tex]Solving for these two equations, and substituting for z in the equation of sphere, we get a circle of radius [tex]2\sqrt{2}[/tex] units.The equation of intersecting circle is:[tex]x^{2} +y^{2}=8[/tex]Now, according to question, parametrizing this region of circle using parameters u and v. Consider cylindrical co-ordinates: (r,θ,z)In cylindrical co-ordinates (x, y, z)= (r cos(θ),  r sin(θ), z)[tex]x^{2} +y^{2}= r^{2}[/tex]Eliminating z, and changing (r, θ)=(u,v)For cone: x=ucos(v)y= usin(v) z=[tex]\sqrt{16-u^{2}[/tex] or (x,y,z)=(ucos(v), usin(v), [tex]\sqrt{16-u^{2}[/tex])where [tex]0\leq u\leq 2\sqrt{2}[/tex] and [tex]0\leq v\leq 2\pi[/tex]