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In spherical coordinates, a sphere can be described as $S = \left\{\theta \in [0,2\pi], \phi \in [0,\pi], r\in [0, R]\right\}$

by letting $x = r\sin \pi \cos \theta, ~ y= r \sin \phi \sin \theta,$ and $z= r\cos \phi$ in the equation $x^2+y^2+z^2 = R$.

This apparently comes from the parametrisation after considering the top surface and bottom surface by drawing a picture. I've tried, but I'm honestly incapable of thinking geometrically.

Question 1: Could someone explain how one can come to this representation algebraically?

Question 2: If we consider the region between the sphere $x^2+y^2+z^2 = R$ and the cone $z = \sqrt{x^2+y^2}$, the upper bound for $\phi$ changes and everything else stays the same: we have

$$S'=\left\{\theta \in [0,2\pi], \phi \in [0,\pi/4], r\in [0, R]\right\}$$

Why is that? I'm thinking because we're only considering a quarter of the sphere, but I'm not sure.

J. Lanky
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1 Answers1

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Let's first describe the outer surface of the sphere: pick a point $P$ on the surface and consider the radius connecting the origin to $P$. Now let $\phi$ be the angle between the radius and the positive $z$ semi-axis: it is evident that the height of $P$ can be written as $r\cos\phi$, where $r$ is the radius of the sphere.

Now consider the circumference obtained by cutting the sphere orthogonally to the $z$ axis and containing $P$. The radius connecting $P$ to the center of the circumference (that is, the distance between $P$ and the $z$ axis) has length $R_C=r\sin\phi$: you can then describe the circumference on coordinates $x$ and $y$ in the usual way: $y=R_C\sin\theta=r\sin\phi\sin\theta$ and $x=R_C\cos\theta=r\sin\phi\cos\theta$, with $\theta \in [0, 2\pi]$. Since we need to sweep these circumferences only from top to bottom and not all the way around, we take $\phi \in [0, \pi]$.

In this way we have described the outer surface of the sphere, but by varying $r$ we can again sweep all possible spherical surfaces with radius between $0$ and $R$, touching all the internal points of the sphere. Hence, $r\in[0, R]$.

I think that now, reasoning on the function of $\phi$, you can better understand the problem in Question 2.

P.S. Sometimes a Google image search is your friend. Spherical coordinates

Michele De Pascalis
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  • Many thanks for the reply. I've seen this image before. My question #1 was how this is achieved algebraically (i.e. without geometry). Do tell me if this request is misguided. For example, in an exam where I've to do question #2, do I've to draw that picture and go through this reasoning all over? Also, I don't understand the sentence "Since we need to sweep these circumferences only from top to bottom and not all the way around, we take..." Why is this sufficient? Why not all the way around? – J. Lanky Apr 08 '18 at 21:11
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    Imagine P traveling along the meridian and the associated circumference sweeping the surface: you will see that the sphere is covered twice. As for number two, as far as my university is concerned, a sound geometrical reasoning justifies the boundaries, but I'm sure you can pull some trigonometry equivalence and justify it algebraically. – Michele De Pascalis Apr 08 '18 at 21:26