# Parametrize the portion of the sphere of radius 4 centred at the origin that lies inside a cylinder above the xy-plane

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, $$0\le\phi\le\frac{\pi}3$$.
• In the answer, however, it is given as $\phi\in[0,\pi/3]$ Sep 18, 2023 at 5:11
• Sorry, got my notation wrong. Usually, $\phi$ is represented as the downward sweep from the $z$-axis, in which case $0\le\phi\le\pi/3$ is correct, but I have represented it as the upward sweep from the $z$-axis. I will edit the question later today to reflect this. Sep 20, 2023 at 5:28