Why Are the Limits Different for Phi and Theta in Spherical Coordinates?

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In spherical coordinates, the limits for phi (0 to 2π) and theta (0 to π) differ due to the geometric representation of angles. Phi represents the azimuthal angle around the z-axis, allowing for a full rotation, while theta represents the polar angle from the z-axis, covering only half the sphere. This distinction ensures that each point in the spherical volume is covered without duplication. The integration process benefits from this symmetry, simplifying calculations. Understanding these limits is crucial for accurately performing volume integrals in physics and mathematics.
Davio
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Hey guys, could one of you explain why when doing a volume integral using spherical polar coordinates, you have the limits as 2 pi to 0 on phi but only pi to 0 on theta? Thanks.

To clarify, I've been doing this all this time for questions, but it just occurred to me that I Don't know why i do that :-p.
 
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Hi Davio! :smile:

(Of course, you can't go from o to 2π on both, because then you'd be covering every point twice!)

Well, you can always do it the other way round … but the integrals that usually occur in practice just happen to be symmetric in phi, so integrating from 0 to 2π on phi is dead easy! :biggrin:

In particular, if you're converting from (x,y,z), then you get ∫∫∫(r^2)sinthetadrdthetadphi … and that itself is symmetric in phi, so with luck the whole thing immediately becomes 2π∫∫r^2)sinthetadrdtheta. :smile:
 
Hmmm, but wouldn't it be, for both phi and theta, just pi to zero - ie. half each? Or am imagining this wrong :-p. R is radius, so integrating along that place, in a circle, theta and phi are both angles, just in different directions? My maths is a bit poor (not good for a physics major)!
 
Hi, Davio!

Take a line segment from the centre of the ball to its surface. That line segment makes the angle theta with the "z"-axis.

In order to reach ALL points on the circle with the same angle to the "xy"-plane, you rotate the line segment around the z-axis, with phi then going from 0 to 2pi.

Now, you will have covered ALL these circles (and hence all points) for theta-values going from 0 (i.e, the line segment runs along the positive z-half axis) to pi, (i.e along the negative half axis for z)
 
… don't ignore America …

Yup, arildno is right!

Start with a very long string, fixed at the North pole. Take the other end down the Greenwich meridian from theta = 0 to π. Now you're at the South pole.

So far so good … :smile:

Now sweep the string round from phi = 0 to π. You'll cover most of Europe and Africa and the whole of Asia and Australia.

But you'll stop at the International Date Line!

:rolleyes: What about America? :rolleyes:
 
One simple way to see it. Latitude goes from -90 to 90, while longitude goes from -180 to 180.
 
It's those blasted physicists again! They keep swapping \theta and \phi on us!
 
I'm going to sit down and think about the replies in a minute, I'm on to a question about conical cones now, does anyone have any good resources for understand the images behind integration? I can integrate etc, but can't quite understand the limits of weird shapes, or even normal shapes!
 
… painting problem …

Hi Davio! :smile:

Do you mean comical cones?

Now, they are weird! :biggrin:

But keep this in perspective … this isn't an integration problem … it's only a painting problem.

Imagine you have to program a robot to paint a sphere … do you tell it to paint from 0 to π, or 2π, for each coordinate?

If you give it the wrong instructions, it'll either waste paint or not use enough! :frown:

That's your only problem … making sure that everything is covered which should be! :smile:
 

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