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Proof of Volume of a Sphere

  1. Feb 12, 2005 #1
    I'm trying to prove the volume of a sphere is (4/3)(pi)r^3. (Sorry I haven't figured out the tex thing yet)

    I was thinking that the volume of a sphere is the sum of the circular cross-sections that make it up. Since "r" is different for each cross-section, you put in the variable "x" and get:

    (pi)x^2.

    The height of the sphere can be represented by the change in y, (dy) so now we get the integral:

    int{ (pi)x^2*dy }

    Since we need the variable of integration in terms of y, I went to the equation of a circle.

    x^2 + y^2 = r^2

    So, x^2 = r^2 - y^2

    Substituting that into the integral we get the final intergral:

    int { (pi)(r^2-y^2)*dy }

    This is as far as I can think this out. Where did I make a mistake or where do I need to go from here?


    -------------
    Jameson
     
    Last edited: Feb 12, 2005
  2. jcsd
  3. Feb 12, 2005 #2

    dextercioby

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    Do you know triple integration and change of variables in triple/multiple integrations...?

    Daniel.
     
  4. Feb 12, 2005 #3
    No I do not. Is that the only way to prove this?
     
  5. Feb 12, 2005 #4

    dextercioby

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    BTW,in case you didn't know,the volume of a sphere is ZERO...

    Daniel.
     
  6. Feb 12, 2005 #5
    Look. If you want to prove how smart you are, do it somewhere else. If you just don't know how to do this problem, then don't say anything. If you do, I would like your help.

    How is the volume of a sphere zero?
     
  7. Feb 12, 2005 #6
    Your method is suitable, buth have you ever heard of spherical polar coordinates?

    It is a radial coordinate r, and two polar angles theta and phi. Notice how this coordinate is a much more natural one, for say the motion of you elbow and fore arm. There is some fun geometry to figure out, but here is an appropriate transformation :

    x = r sin ( phi) cos ( theta)

    y = r sin ( phi )sin (theta)

    z = r cos ( phi )

    using this, you can transform dx, dy, and dz in to dr etc

    Then just do a triple integral where r goes from 0 to R, the radius of the sphere. Theta goes from 0 to 2pi and Phi goes from zero to pi. My explanation is not so great, but you should look up spherical polar coordinates.
     
  8. Feb 12, 2005 #7

    dextercioby

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    I'm saying the volume of a sphere is ZERO,USING THE DEFINITION OF A SPHERE.

    As for your initial method,it was incorrect.You were insuccessfully attempting to find the volume of a ball of radius R using cylindrical coordinates...

    Daniel.
     
  9. Feb 12, 2005 #8

    dextercioby

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  10. Feb 12, 2005 #9

    cepheid

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    Dextercioby always tells people that what they have said is wrong, based on semantics. That makes it impossible for them to understand why. All it does it confuse the issue. It's not a very good way to confront people who are trying to learn IMHO. Dexter is pointing out that a sphere is a two-dimensional surface in 3D space, therefore it has no volume. However, the volume **ENCLOSED** by that spherical surface is given by [itex] \frac{4}{3} \pi r^3 [/itex]. Arrrgggh! See what I mean? Frustrating guy! :cry: The volume of a spherical solid is given by that formula too.

    You CAN use single variable calculus to determine that this formula is correct, i.e. you don't have to perform a triple integration, only a single one. Say you have a sphere centred at the origin. One of those thin (let's make them vertical) slices you were talking about has area pi r^2 (presumably you already derived the formula for the area of a circle. If not, you can use calculus for that too). The slice has an infinitesimal thickness dx. And as you move along the x axis from -r to r, through the sphere, you're encountering larger and larger slices, with the largest being at the centre, and then smaller and smaller, till you get to the other end. You want to integrate all of these "infinitesimal" volumes. But the radius of each slice is not constant! It is a variable. If you are a distance x away from the centre of the sphere, the radius of your slice is 'y', where y is the y coordinate of the point on the sphere at that x coordinate. Using pythagoras, [itex] y = \sqrt{r^2 - x^2} [/itex]

    Integrate:

    [tex] V = \int_{-r}^{r} {\pi (r^2 - x^2)dx} = 2 \pi \int_0^r {(r^2 - x^2)dx} [/tex]

    EDIT: LOL...i didn't need to explain any of this to you. You had already derived it in your first post! So, have you encountered some specific problem in trying to integrate that?
     
    Last edited: Feb 12, 2005
  11. Feb 12, 2005 #10
    Thank you very much for your explaination. I don't know how to integrate [tex] 2 \pi \int_0^r {(r^2 - x^2)dx} [/tex] since it has two different variables, but I can find out how to do that.

    Many thanks,
    Jameson
     
    Last edited: Feb 13, 2005
  12. Feb 12, 2005 #11

    cepheid

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    r is not a variable. It is the radius of your sphere, hence it is constant. I hope that helps you out!
     
  13. Feb 12, 2005 #12

    dextercioby

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    It's not SEMANTICS,it's plain (simple) mathematics (which 'encloses' a great deal of logics)...The volume in question EXISTS,but it's zero...

    Daniel.
     
  14. Feb 13, 2005 #13

    Curious3141

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    Jameson, calculus is the "easy" way to do it (that is, if you're an advanced student. :biggrin: ) but I'm guessing you'll love this non-calculus method to the same end :

    http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html

    This method was originally due to Archimedes. Cool guy. :smile:
     
  15. Feb 13, 2005 #14

    Gokul43201

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    I don't see how. It looks perfectly good to me.
     
  16. Feb 13, 2005 #15

    saltydog

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    Me neither

    Yea, I know I'm slow. Anyway, here's the volume using a triple integral. And I didn't know either what Daniel meant about the volume being zero, and in fact it took me a while to figure it out even after Cepheid explained it.

    In spherical coordinates, the problem can be defined as follows:

    [tex] vol=8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\int_0^r \rho^2 \sin(\phi)d\rho d\theta d\phi [/tex]

    Beautiful isn't it!

    So:

    [tex] 8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2} \sin(\phi)(\frac{\rho^3}{3}){|}_0^r d\theta d\phi [/tex]

    and then:

    [tex] \frac{4r^3 \pi}{3}\int_0^\frac{\pi}{2}sin(\phi)d\phi [/tex]

    or:

    [tex] -\frac{4r^3\pi}{3}[0-1]=\frac{4\pi r^3}{3}[/tex]

    Don't you just love Calculus!
     
  17. Feb 13, 2005 #16
    Thanks to everyone for their help. I still don't quite understand how "r" is a constant in the integral... that would mean you could pull it out in front of the equation and get:

    2(pi)r^2 * int { -y^2 dy }

    Is that the correct answer?

    Thanks again to all,
    Jameson
     
  18. Feb 13, 2005 #17

    dextercioby

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    No.You'll have to integrate "r^{2}" like any other constant.So if you pull it out of the integral,u must multiply with the length on the interval:

    [tex] C\int_{a}^{b} dy=C(b-a) [/tex]

    Daniel.
     
  19. Feb 13, 2005 #18

    Galileo

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    The radius of the sphere is a given for the problem. It does not depend on x or y.

    [tex] 2 \pi \int_0^r {(r^2 - x^2)dx}=2\pi\left(\int_0^r r^2dx-\int_0^rx^2dx\right)[/tex]
    Then you can take the [itex]r^2[/itex] out of the first integral.
     
  20. Feb 13, 2005 #19
    i think one could prove it without calculus.
     
  21. Feb 13, 2005 #20

    mathwonk

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    jameson, after reading your impatient outbursts at people for rudeness, i do not feel like helping you. when you post on a public forum like this, anyone at all may respond. in my opinion, you are asking for something for nothing, and should be grateful for whatever you get of value.
     
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