Question on spherical integration

  • #1
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11

Homework Statement


So if you integrate over a spherical area, say a ball of radius 1, then 0≤p≤1, 0≤θ≤2∏, and 0≤∅≤∏. My question is why don't you integrate ∅ between 0 and 2∏? I mean if you are integrating over a sphere then you have to go around it vertically AND horizontally completely? wouldn't both the angles be from 0 to 2∏?

Thanks genius's :D

edit: does it have something to do with the fact that you add a p^2sin(phi) when integrating in spherical coordinates?
 

Answers and Replies

  • #2
35,292
7,149

Homework Statement


So if you integrate over a spherical area, say a ball of radius 1, then 0≤p≤1, 0≤θ≤2∏, and 0≤∅≤∏. My question is why don't you integrate ∅ between 0 and 2∏? I mean if you are integrating over a sphere then you have to go around it vertically AND horizontally completely? wouldn't both the angles be from 0 to 2∏?

Thanks genius's :D

edit: does it have something to do with the fact that you add a p^2sin(phi) when integrating in spherical coordinates?
No, it's much simpler than that.

A point (r, θ) describes any point in the horizontal plane, and as you know, 0 ≤ θ ≤ ##2\pi##. To describe a point in space, and we can do this by adding a 3rd coordinate, ##\phi##. All we need to identify this point in space is a direction angle, ##\phi##. Keep in mind that ##\phi## is measured from the positive z-axis, so having a range of [0, ##\pi##] gets us all the way from straight up to straight down.

Note that r is the distance from the origin to the point in polar and cylindrical coordinates. In spherical coordinates, the distance from the origin to the point is ρ, "rho."
 

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