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Integrating 2*pi*R and pi*R^2

  1. May 3, 2005 #1
    i noticed that if i integrate
    [tex]2 \pi r[/tex]
    i get
    [tex]\int 2 \pi r dr=\pi r^2[/tex]
    i figured its because the area of a circle can be seen as the sum of circumference's of circles with radius 0 to radius [tex]r[/tex]
    i was thinking if the half volume of a ball also be seen as made from the sum of areas of circules with radius 0 to radius [tex]r[/tex]?
    integrating [tex]\pi r^2[/tex] gives [tex]\pi \frac{r^3}{3}[/tex]
    so multiplying by two should give the ball volume formula but it does not why is that?
  2. jcsd
  3. May 3, 2005 #2
    The the area integral is a double integral over r and the azimuthal angle. The volume integral is over r, the azimuthal angle and the polar angle. You're missing some parts of the "big picture" about spherical coordinates.

    You kind of skipped the intergrating over the azimuthal angle part by including [tex] 2 \pi [/tex] straight into your integrand. Check http://mathworld.wolfram.com/Sphere.html and http://mathworld.wolfram.com/SphericalCoordinates.html for details.
    Last edited by a moderator: Apr 21, 2017
  4. May 3, 2005 #3


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    In other words, the derivative of the area of a circle, [tex]\pi r^2[/tex] ,with respect to r, is the circumference of the circle [tex]2\pi r[/tex]!

    Your formula for the volume of a sphere is incorrect. The volume of a sphere is [tex]\frac{4}{3}\pi r^3[/tex] and its derivative is indeed the surface area of the sphere: [tex]4\pi r^2[/tex]
  5. May 3, 2005 #4
    alright then i have another question
    would it be correct to define a half of a ball with diameter D as the set of all circles whose diameters range from 0 to D?
  6. May 3, 2005 #5


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    No.That would be a hemisphere.A half of a ball would be a continuous array of disks whose centers lie on a line segment of length R (radius of the ball).

  7. May 3, 2005 #6
    aren't a hemi sphere and a half ball the same things?
    maybe i didn't form the question correctly by circles i mean circles with their area not just the circumference i suppose one could call it a disk.
    ill reform the question:
    can a half ball with diameter D be defined as a set of all discs with diameter's ranging from 0 to D and parallel to each other.
  8. May 3, 2005 #7


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    One calls that a disk...

    Yes,but it's highly reccomendable to define it as a half of a ball. :tongue2:

  9. May 3, 2005 #8
    im not just asking these questions for nothing im trying to get to a point :smile:
    alright then if a half of a ball can be defiend this way and [tex]A(D)= \pi (\frac{D}{2})^2[/tex] is a function which gives the area of a disk of diameter D then shouldn't its integral [tex]\int A(D)[/tex] which is actually the sum of all the areas of all the discs with diameters ranging from 0 to D which make the half ball give the volume formula of a half ball of diameter D?
  10. May 3, 2005 #9


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    It does,doesn't it...?

    [tex]\int_{0}^{D} \pi\frac{D^{2}}{4} \ dD=\pi\frac{D^{3}}{12}=\frac{2\pi R^{3}}{3}=V_{\mbox{half ball}} [/tex]

  11. May 3, 2005 #10
    o right i didnt notice that.
    how come if you integrate [tex]\pi R^2[/tex] it doesn't come out right whats the diffrence between integrating by the diameter or by the radius?
  12. May 3, 2005 #11


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    You've been explained that u should normally integrate the area of the surface (sphere) enclosing the volume (ball) to get it right.It works for sphere-ball and their 1D-2D counterparts:circle-disk...It doesn't for cube surface-cube and neither for square contour-square...

  13. May 3, 2005 #12
    i understand that integrating the surface will work but im trying to understand why integrating the area function of a disk by diameter gives the correct formula for the volume of a half of a ball but integrating by the radius doesn't even though the idea is the same (adding up the areas of the discs that make the half ball).
  14. May 3, 2005 #13
    Because when you set up the equation for the integral according to the radii, you did not take into account the curvature of the ball... The radius of the disks do not have a linear relationship with the distance from the center (or from the top). You derived a formula of a cone, which is also a bunch of discs centered on the same axis. The difference between the cone and the half-ball is that the half ball bulges out like a circle, while the cone stays linear.
  15. May 3, 2005 #14
    It's because you didn't the change of variable correctly : D=2r->dD=2dr
    boundary : D_0=2R

    [tex] \int_0^{D_0}\pi D^2/4dD=\int_0^{2R}2\pi r^2dR=2/3\pi r^3 [/tex]
  16. May 4, 2005 #15
    It is just some 'curiosity' that this actually works:
    "The sum of all the areas of all the discs with diameters ranging from 0 to D" gives the volume of half a ball

    [tex]\int_{0}^{D} \pi\frac{D^{2}}{4} \ dD=\frac{1}{12}\pi D^3=\frac{2}{3}\pi R^3=V_{\mbox{half ball}} [/tex]

    or similarly,

    [tex]\int_{0}^{2R} \pi R^2 \ dR=\frac{2}{3}\pi R^3=V_{\mbox{half ball}} [/tex]

    but this is not based on proper reasoning and it does not take the curvature of the ball into account (As Moo of Doom seems to suggest) it is the volume of a cone with base radius R and height 2R (which is the same as the volume of half a ball with radius R)

    "The sum of all the areas of all the discs with radi ranging from 0 to R" only gives the volume of a quart of a ball (i.e. of a cone with base radius R and height R)

    [tex]\int_{0}^{R} \pi R^2 \ dR=\frac{1}{3}\pi R^{3}=V_{\mbox{quart ball}} [/tex]
  17. May 5, 2005 #16
    I suggest to go through (Centroid)Theorems of Pappus.
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