Question

Find a parametric representation for the surface. The part of the sphere x2 + y2 + z2 = 144 that lies between the planes z = â’6 and z = 6. (Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of θ and/or Ď•.) (where â’6 < z < 6)

Answer 1

We can use the standard set of parametric equations in spherical coordinates:

$\begin{cases}x(u,v)=12\cos u\sin v\\y(u,v)=12\sin u\sin v\\z(u,v)=12\cos v\end{cases}$

with $0\le u\le2\pi$. Then the part of the sphere that lies between the planes $z=-6$ and $z=6$ is determined by the domain of $v$. To find that, we can sketch a picture of the situtation (attached below).

The right triangle in the image has its hypotenuse coinciding with the radius of the sphere so its length is 12, and the vertical leg has length 6 because it coincides with the plane $z=6$. The angle between them is the upper limit of $v$. At any point along the intersection of the sphere with the plane $z=6$, we want to have

$\cos v=\dfrac6{12}=\dfrac12\implies v=\dfrac\pi3$

and so by symmetry, we need to set $-\dfrac\pi3\le v\le\dfrac\pi3$