Confused about the intergral over a sphere

  • Thread starter khkwang
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  • #1
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I don't know the beginning part of the question is relevant, so I'll leave it out unless requested.

At the point of:

[tex]\int[/tex]sin[tex]\vartheta[/tex]d[tex]\vartheta[/tex]d[tex]\phi[/tex]

Which is to be integrated over a sphere, when integrating from 0 to pi for [tex]\vartheta[/tex] and then from 0 to 2pi for [tex]\phi[/tex], we get 4pi, which is the answer I'm looking for.

But if I integrate [tex]\vartheta[/tex] from 0 to 2pi first, then [tex]\phi[/tex] from 0 to pi, I get 0, which is definitely not the answer I'm looking for.

Can someone tell me why this is? And how I know which to integrate over pi and which to integrate over 2pi generally?
 

Answers and Replies

  • #2
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Hi! The sin(theta) comes from the way the angular coordinates are defined. So you cannot simply say that you will integrate theta from 0 to 2pi. You can check out spherical cooordinates on wikipedia and the derivation of the integral measure.

Here is a way to remember: The coordinate the goes from 0 til pi is the one in the sin() function in the integration measure, because otherwise sin() would become negative. This integral measures the area on the sphere, and it should not be possible to get a negative answer, since area is positive.

Of course, each time someone uses a spherical coordite system, they should say how it is defined.

Torquil
 
  • #3
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Ah, I'm not sure I understand all that (the first part), but I understand the part about the area not being possible to be negative. Thanks.
 
  • #4
HallsofIvy
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I don't know the beginning part of the question is relevant, so I'll leave it out unless requested.

At the point of:

[tex]\int[/tex]sin[tex]\vartheta[/tex]d[tex]\vartheta[/tex]d[tex]\phi[/tex]

Which is to be integrated over a sphere, when integrating from 0 to pi for [tex]\vartheta[/tex] and then from 0 to 2pi for [tex]\phi[/tex], we get 4pi, which is the answer I'm looking for.

But if I integrate [tex]\vartheta[/tex] from 0 to 2pi first, then [tex]\phi[/tex] from 0 to pi, I get 0, which is definitely not the answer I'm looking for.

Can someone tell me why this is? And how I know which to integrate over pi and which to integrate over 2pi generally?
Unfortunately, to confuse the situation further, mathematics and engineering use opposite conventiions. Mathematics uses [itex]\theta[/itex] as the "longitude" and [itex]\phi[/itex] as the "co-latitude". Engineering reverses those meanings.

khkwang is using "engineering" notation, but either way, longitude goes from "180 degrees east to 180 degrees west" or a full 360 degree, [itex]2\pi[/itex] radian, circle while latitude only goes from "90 degrees south to 90 degrees west", only a difference of 180 degrees or [itex]\pi[/itex] radians.
 
  • #5
Landau
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Also see the discussion at Wolfram about spherical coordinates.
 
  • #6
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Unfortunately, to confuse the situation further, mathematics and engineering use opposite conventiions. Mathematics uses [itex]\theta[/itex] as the "longitude" and [itex]\phi[/itex] as the "co-latitude". Engineering reverses those meanings

Because of that, I use [itex]\zeta[/itex] for "zenith angle" (co-latitude, the angle from the Cartesian z-axis, [0,[itex]\pi[/itex]]) and [itex]\alpha[/itex] for "azimuth angle" (longitude, the angle from the Cartesian x-axis, [0,2[itex]\pi[/itex])).

Have any notable disasters been attributed to this difference in convention?
 
  • #7
HallsofIvy
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Since, fortunately, it is only engineers who design things, not mathematicians, no!
 

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