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Homework Help: Integration bounds when integrating a half sphere

  1. Jan 31, 2007 #1
    I have to integrate over the northern hemisphere of a sphere. The answer states that the integration bounds for r is 0 to R, for theta it's 0 to pi/2, for phi it's 0 to 2pi.

    What I don't understand is why theta doesn't go from 0 to 2pi. If I had to set this up, I would have theta going from 0 to 2pi and phi going from 0 to pi.

    Can someone please explain this.
  2. jcsd
  3. Jan 31, 2007 #2


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    I'll assume you mean integrating over the volume of the northern hemisphere, since you say r varies from 0 to R, and is not fixed at R (for a surface integration).

    Phi is the angle in the plane of the equator, and theta is the angle down from the North Pole. So when you sweep theta pi/2, that's the half of a longitude line that lies in the northern hemisphere. When you sweep that vertical longitude line all the way around for phi = 0 to 2pi to cover the full northern hemisphere.

    If instead you sweep theta from 0 to 2pi, that is the full length of the longitude line from North Pole to South Pole.
  4. Feb 1, 2007 #3
    What happens if you do [tex] \frac {dx}{dr} of \frac {2}{3} \pi r^3?[/tex] or the volume of half a sphere, what do you end up with?:smile:

    [tex]2 \pi r^2[/tex]


    Last edited: Feb 1, 2007
  5. Feb 1, 2007 #4


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    In their never ending battle to confuse one another, mathematicians and physicists use opposite conventions for [itex]\theta[/itex] and [itex]\phi[/itex]. In most calculus books, you will see [itex]\theta[/itex] used for the "longitude" (measured around the equator) and [itex]\phi[/itex] for the "co-latitude" (measured down from the north pole)- physics text books reverse that. You are using the "physics" convention. In this case, as you have been told, the measurement around the equator, [itex]\phi[/itex], for the entire sphere, goes all the way around- from 0 to [itex]2\pi[/itex]. Notice that if the co-latitude, [itex]\theta[/itex], also went from 0 to [itex]2\pi[/itex], points would have two sets of coordinates. The point with [itex]\phi= 3\pi/2[/itex], [itex]\theta= \pi/2[/itex] could also be labeld [itex]\phi= \pi/2[/itex], [itex]\theta= 3\pi/2[/itex]. Taking [itex]\theta[/itex] from 0 to [itex]\pi[/itex] only solves that problem. Of course, to get the "northern hemisphere" only, you take [itex]\theta[/itex] from 0 to [itex]\pi/2[/itex].
    Last edited by a moderator: Feb 2, 2007
  6. Feb 1, 2007 #5
    You'll also notice if you multiply [tex] \frac {3\pi}{2}. \frac{\pi}{2}= \frac {3}{4} \pi^2[/tex]

    Try integrating that with the radius? What do you end up with? yup

    [tex] \frac {4}{3} \pi r^3[/tex]

    [tex] \frac{dy}{dx}\frac {4}{3}\pi r^3=[/tex]

    :smile: It all makes beautiful sense, and it's where Archimedes got his calculations from, although his integration methods would have been different from the way modern integration worked, and he had no general rules to work from :smile:

    EDIT: neat way of checking your answers too :smile:
    Last edited: Feb 1, 2007
  7. Feb 1, 2007 #6


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    Wow, thanks!! That really clears things up. It's funny, when I pulled out a textbook at home last night to double-check the convention so that I could reply to the OP, I pulled out a physics text first, and used that convention (which apparently was the same as the OP's problem was assuming). But as I looked at the convention in the text, I was saying to myself how I thought it was just the opposite! I feel better now. :blushing:
  8. Feb 1, 2007 #7
    What I don't get is this:

    When integrating over an entire sphere, [tex]\theta[/tex] would go from 0 to [tex]2\pi[/tex]. So if I integrate a half sphere, why not go from 0 to [tex]\pi[/tex]? That would cover the range of [tex]\theta[/tex] that is in the northern hemisphere.

    EDIT: Never mind, I think I can see it now. So if I was integrating the entire sphere, [tex]\theta[/tex] would just go from 0 to [tex]\pi[/tex] right?
    Last edited: Feb 1, 2007
  9. Feb 1, 2007 #8


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    Correct. That would pick up the southern hemisphere as well.
  10. Feb 1, 2007 #9
    Well I was useless as usual but, I'm used to it. what is 1/2 of. Check your answers with the simple formula that = dy/dx. It's just so obvious if you look at what lies behind it. I guess I tried to convey that and failed.
  11. Feb 2, 2007 #10


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    How did the [itex]\pi^2[/itex] suddenly become [itex]\pi[/itex]

  12. Feb 2, 2007 #11
    it didn't but I assume if you put r into the equation or relate it, the formula becomes :-

    [tex]\int \frac{3}{4} \pi^2=\frac{4}{3} \pi^3+C[/tex]

    Accounting for the radius in relation to a sphere=

    [tex]\frac{4}{3}\pi r^3 [/tex]

    Am I wrong?
    Last edited: Feb 2, 2007
  13. Feb 2, 2007 #12


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    [tex]\int \frac{3}{4} \pi^2.=\frac{4}{3} \pi^3+C[/tex]

    What does the . represent there? What are you integrating with respect? HOW are you putting r in there?
  14. Feb 2, 2007 #13
    Oh it's a typo I just mean r in respect to a sphere given that area always integrate to a..

    OK I edited it. now theirs no . :smile:

    It's all covered in the article I quoted anyway.


    [tex](x - x_0 )^2 + (y - y_0 )^2 + ( z - z_0 )^2 = r^2 \,.[/tex]

    [tex]x = x_0 + r \sin \theta \; \cos \phi
    y = y_0 + r \sin \theta \; \sin \phi \qquad (0 \leq \theta \leq \pi \mbox{ and } -\pi < \phi \leq \pi) \,
    z = z_0 + r \cos \theta \,)[/tex]


    [tex]A = 4 \pi r^2 \,[/tex]

    [tex]V = \frac{4}{3}\pi r^3.[/tex]
    Last edited: Feb 3, 2007
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