Parametrize the portion of the surface of the sphere of radius 4 centred at the origin that lies inside the cylinder determined by $x^2 + y^2 = 12$ and above the xy-plane.
Solution: The radius of the sphere and cylinder are 4 and $2\sqrt{3}$. Using the spherical coordinates, a parametrization is , $$\vec G(\phi, \theta) =(4 sin\, \phi \,\,cos \,\theta, 4 sin \,\phi\,\, sin \theta, 4 cos \,\phi)$$ where $\theta\in [0,2\pi]$. Please help me find the range of $\phi$.
Pick a point on the intersection of the sphere and cylinder. Call this point $w$. Let $A$ be the line from $w$ to the origin. Let $B$ be the vertical line contained by the cylinder from $w$ to $z=0$. Let $C$ be the line from the origin to the endpoint of $B$ at $z=0$. Then, we get the following right triangle,
Now, $B$ is a line in the surface of the cylinder, and the curve in the diagram is the surface of the sphere in the triangle's plane. Thus, $A$ must be the radius of the sphere, and $C$ must be the radius of the cylinder. Furthermore, the angle $\psi = \frac{\pi}2-\phi$.
Thus $|A|=4$ and $|C|=2\sqrt{3}$. Finding the angle $\gamma$, we expect that $\sin(\gamma) = \frac{|C|}{|A|}=\frac{2\sqrt{3}}4=\frac{\sqrt{3}}2$. Of course, $\sin(\gamma)=\frac{\sqrt{3}}2$ when $\gamma = \frac{\pi}3$. Since $ABC$ is a right angle triangle, then $\delta = \frac{\pi}2$. Thus, $\gamma+\delta+\psi = \frac{\pi}3+\frac{\pi}2+\psi = \pi$. So, $\psi=\frac{\pi}6$. Finally, $\psi = \frac{\pi}6= \frac{\pi}2-\phi$, so $\phi = \frac{\pi}3$.
Now, consider the radius of the sphere sweeping down from the $z$-axis toward the line $A$. As the radius sweeps downward, $\phi$ increases from $0$ to $\frac{\pi}3$. Thus, $0\le\phi\le\frac{\pi}3$.