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For something like $(x-1)^2+y^2+z^2=1$, we would let

$x-1=\rho \sin (\phi)\cos (\theta)$

$y=\rho \sin (\phi)\sin (\theta)$

$z=\rho \cos (\phi)$

But we know $\rho =1$ right? So it becomes

$x-1= \sin (\phi)\cos (\theta)$

$y= \sin (\phi)\sin (\theta)$

$z= \cos (\phi)$

If the sphere was centred on the origin, then the limits would be $\phi \in [0, \pi]$ and $\theta \in [0, 2\pi]$ but since its not, I am guessing it isn't this. I need to know how to find the limits when it is not in the centre.

I know that the $\phi$ is the angle from the positive $z$ axis downwards but it is strange because if the sphere is in the origin, and if the limit is from $0$ to $\pi$, it doesn't cover all of it, it covers half. Is it always meant to cover half?

Also with $\theta$, is it the angle on the $xy$ plane with respect to the origin or on the $xy$ plane with respect to the centre of the sphere?

snowman
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  • When you made the left-hand side of your equations $x-1$, $y$, and $z$, you shifted your spherical coordinates. Meaning, the $z$ axis becomes shifted to vertical line over $(1,0,0)$. If this is your intention, then you would have the usual limits for $\phi$ and $\theta$, except you interpret $\phi$ as the angle from the top half of your new line. – user197427 Apr 27 '15 at 19:13
  • ohh so everything is with respect to the centre? and the limits would remain the same is what you are saying correct? – snowman Apr 27 '15 at 19:15
  • It really depends on what you want. If you do not mind that $\phi$ is measured from a different line, then yes. If you must have $\phi$ as an angle from the positive $z$ axis ("true" spherical coordinates), then the situation is a bit more complicated. – user197427 Apr 27 '15 at 19:19
  • OK so do you recommend me to just measure from the shifted z axis to the centre? Also I really don't understand why phi just measures half of the sphere and not all of it... please would you explain this. – snowman Apr 27 '15 at 19:21

1 Answers1

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Some comments first. For the usual spherical coordinates:

(1) $\theta$ is measured as the angle with respect to the origin in the $xy$ plane. On the unit sphere, when you let $\theta$ be a constant value, e.g. $\theta = \frac{\pi}{4}$, then you will get a curve that is a longitude of the sphere. This is half of a circle joining the north and south pole.

(2) Letting $\phi \in [0, \pi]$ does indeed cover the whole sphere. On the unit sphere, when you set $\phi$ to be a constant value, then you will get a curve that is a latitude of the sphere. These are horizontal circles. You can imagine these by taking a point in the $xz$-plane with a certain $\phi$ angle and then rotating the point around the $z$-axis. The value of $\phi$ gives the "height" of the circle. For instance, the north pole is $\phi = 0$, the equator is $\phi = \frac{\pi}{2}$, and the south pole is $\phi = \pi$, with various intermediate values giving intermediate circles.

Now, if you want to use the usual spherical coordinates to describe your sphere, then you should not shift the coordinate system. Instead, just expand: $$x^2 - 2x + 1 + y^2+z^2 = 1.$$ So, $$x^2+y^2+z^2 = 2x, $$ which becomes $\rho^2 = 2\rho \sin \phi \cos \theta$, using $x = \rho \sin \phi \cos \theta$, $y = \rho \sin \phi \sin \theta$, $z = \rho \cos \phi$. In other words, the equation is $\rho = 2 \sin \phi \cos \theta$. Substituting into the typical spherical coordinates, we get $$\begin{align*} x & = 2\sin^2 \phi \cos^2\theta \\ y & =2\sin^2 \phi \sin \theta \cos \theta \\ z& = 2 \sin \phi \cos \phi \cos \theta. \end{align*}$$

For the bounds of $\phi$, sketch a picture of your sphere and slice it vertically by a plane containing the $z$-axis (this is a plane with constant $\theta$ value). You will get a circle touching the origin. On this circle, $\phi = 0$ corresponds to the origin. As $\phi$ increases, we move "up" the circle. $\phi = \frac{\pi}{2}$ corresponds to the other point where the circle intersects the $xy$ plane, and $\phi = \pi$ again brings us back to the origin. Since this image doesn't really depend on $\theta$, the values of $\phi$ are $[0,\pi]$.

The bounds for $\theta$ may be slightly harder to see since the the slices with $\phi$ constant are cones are various angles. Intersecting these with the shifted sphere may be tough to imagine, so let us consider the intersection with $\phi = \frac{\pi}{2}$. This is the intersection with the $xy$-plane, giving a circle of radius 1 centered at (1,0). $\theta = -\frac{\pi}{2}$ is the origin, then you move counterclockwise around the circle to (2,0) which is given by $\theta = 0$ and then back to the origin at $\theta = \frac{\pi}{2}$. So, $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (this holds in general).

user197427
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