Question

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 1

Answer:

(x,y,z)=(ucos(v), usin(v), $\sqrt{16-u^{2}$)

where $0\leq u\leq 2\sqrt{2}$ and $0\leq v\leq 2\pi$

Step-by-step explanation:

Equation of a cone is $z=\sqrt{x^{2} +y^{2}}$

Equation of a paraboloid is $z=x^{2} +y^{2}$

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

$x^{2} +y^{2} +z^{2} = 16\\z=\sqrt{x^{2} +y^{2}}$

where $z\geq x^{2} +y^{2}$

Solving for these two equations, and substituting for z in the equation of sphere, we get a circle of radius $2\sqrt{2}$ units.

The equation of intersecting circle is:

$x^{2} +y^{2}=8$

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)

$x^{2} +y^{2}= r^{2}$

Eliminating z, and changing (r, θ)=(u,v)

For cone: x=ucos(v)

y= usin(v)

z=$\sqrt{16-u^{2}$

or (x,y,z)=(ucos(v), usin(v), $\sqrt{16-u^{2}$)

where $0\leq u\leq 2\sqrt{2}$ and $0\leq v\leq 2\pi$