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Check volume of sphere

  1. May 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Check that the volume of a sphere is 4/3(pi)r^3
    (use disk method)
    So I don't understand it I got stuck so I looked. I still don't get their solution.
    Chapter 7, section 2, question 59. http://www.calcchat.com/book/Calculus-ETF-5e/


    2. Relevant equations



    3. The attempt at a solution

    So my question is if you look at their solution they integrate from -r to r. I follow this. Then they integrate from 0 to r but they multiply the integral by 2. I'm OK with this also. What I'm not OK with is the diagram. If you look the line y = 0 is the end of the diagram. Now I think to myself how do you account for the area that is below the x axis? I don't quite get it. I just don't understand why it is basically half.
    Thanks, Hope that quesiton makes sense.
     
  2. jcsd
  3. May 25, 2013 #2

    verty

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    That site doesn't work for me but it sounds like you now know enough to answer the question for yourself, by whatever method is most comfortable for you.
     
  4. May 25, 2013 #3
    Well that's great the site doesn't work for you. I still have my question.
    Thanks,
    j
     
  5. May 25, 2013 #4

    SteamKing

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    If you split the sphere into two pieces using the x-z plane, is the volume above the x-z plane the same as the volume below? Doesn't that suggest why the factor of 2 is multiplying the integral?
     
  6. May 25, 2013 #5
    OK, the integrate from 0 to r if you look at the drawing that is in quadrant 1 right? So then then multiply by a factor of 2 which accounts for quadrant 2. But what about 3 and 4? If you were to rotate what they did wouldn't that only be the top half of the sphere? Like their drawing?
    This is what I don't get.
     
  7. May 25, 2013 #6

    SteamKing

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    Let's ignore post #4.

    The radius of the disk located at x is R(x) = sqrt (r^2 - x^2). The area of this disk is pi * R(x)^2 and the
    volume of the disk is pi * R(x)^2 * dx. The original integral to get the volume of the sphere was taken over the interval [-r, r]. However, the function is symmetric about the y-axis, so the limits can be changed to [0,r] and the integral is multiplied by 2 to obtain the same value as the integral with the previous limits of integration.
     
  8. May 26, 2013 #7
    If you are using "discs" you are trying to find the volume of a sphere of radius r by rotating the part of a circle in the first quadrant around the x-axis (y=0) to produce a half sphere. Therefore you have to multiply the volume integral by 2 to get the volume of the full sphere.

    In other words if the equation of the circle is

    [tex] x^{2} \, + \, y^{2} \, = \, r^{2} [/tex]

    you solve for y to get

    [tex] y^{2} = \pm \sqrt{r^{2} \, - \, x^{2}} [/tex]

    and the portion of the circle in the first quadrant given by

    [itex] y \, = \, \sqrt{r^{2} \, - \, x^{2}} [/itex] where [itex]0 \leq x \leq r [/itex]

    So, when you apply the shell method to the quarter circle in the first quadrant from 0 to r what integral do you get ? Remember you have to multiply this integral by 2 otherwise you'll only get the formula for the volume of a half sphere.

    P.S. In the problem you linked to he writes the first integral using the bounds

    [tex] -r \leq x \leq r [/tex]

    if you evaluate the integral from -r to r you'll get a full sphere when you rotate around the x-axis and you don't have to multiply the integral by 2. But in the next step he changes the bounds from 0 to r and when you rotate that around the x-axis you get a half sphere so you have to multiply the integral by 2 in order to get the full volume.

    P.S. If you continue on in your studies of analytic geometry/calculus you'll eventually learn this can also be easily done via multivariable calculus using triple integrals in spherical coordinates. But that is for another time and not for now. ;)
     
    Last edited: May 26, 2013
  9. May 26, 2013 #8

    SteamKing

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    That's one way to do it, but that method is based on the Theorem of Pappus. In the method of disks, as I understand it, the volume is approximated by stacking very short cylindrical segments, or disks, one on top of the other. For the sphere, the disks all have different radii, which are determined based on the position of the disk from the center of the sphere.
     
  10. May 26, 2013 #9
    Thanks Skins
     
  11. May 26, 2013 #10

    dextercioby

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    The volume of a sphere is not defined, but the volume enclosed by a sphere is.
     
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