1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: How do i prove the volume of a sphere by double integration?

  1. Apr 19, 2009 #1
    using the technique of double integration i need to prove that the volume of a sphere =
    4/3*Pi*r^3....i kinda have some idea but dont know what original function to start the double integration on?


    Thank you...
  2. jcsd
  3. Apr 19, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    A couple of alternatives:

    A) Disk method:
    Think of slicing the ball in horizontally placed disks about the vertical, where the vertical variable is called z

    Use the Pythagorean theorem to derive the local radii as a function of z, and integrate the area function for the disks from -r to r.

    B) Shell method:
    Think of the sphere as consisting of concentric spherical SHELLS, with radii varying from 0 to r.
    Find the area function for the shells as a function of the radial variable, and integrate.
  4. Apr 19, 2009 #3
    It says i need to use the multiple integration texhnique....
  5. Apr 19, 2009 #4
    What's the equation for a sphere? Integrate that function around a circular region of some radius, r.
  6. Apr 19, 2009 #5
    I dont know, iv found a few different equations for a sphere and not sure which one to believe...or use! one was x^2+y^2+z^2 = r^2
    another one used sin(theta)...
  7. Apr 19, 2009 #6
    Both of the equations you mentioned can be used.

    Say you have a given sphere in R3, centered at the origin (0,0,0).

    I am going to use spherical coordinates, as it is a lot easier to calculate the volume of a sphere with spherical coordinates.

    Basically, if you integrate teh function 1 over a volume, you get the volume of the object.

    EDIT: thed function you integrate over is 1 in normal coordinate system, but when yoiu go to spherical coordinates you have to multiply the function 1 by the Jacobian, rho^2 sin(phi). So the integral is tripple integral of rho^2 sin(phi), according to the following diagram and limits. you are integrating over Rho, phi, and theta:

    Rho: <0,r>
    Phi: <-pi/2 , pi/2>
    Theta: <0 , 2pi>

    This will yield the volume using a Tripple integral.

    For some reason the Latex code isnt working right now.

    [tex]\int ^{0}_{r} \int ^{-\pi / 2}_{\pi / 2} \int ^{0}_{2\pi} \rho ^{2} sin(\phi ) d\phi d\theta d\rho[/tex]

    for some reason the latex command is showing up wrong, if someone else can post it...
    \int ^{0}_{r} \int ^{-\pi / 2}_{\pi / 2} \int ^{0}_{2\pi} 1 d\phi d\theta d\rho
    Last edited: Apr 19, 2009
  8. Apr 19, 2009 #7
    You can also use this method. Consider the integral of exp(-r^2) over R^3. Clearly this is given by:

    I = Integral from r = 0 to infinity of A r^2 exp(-r^2) dr

    where A r^2 is the area of a sphere of radius r. We want to compute A, and from that the volume of the sphere follows by integrating that area formula, yielding A/3 r^3.

    The integral can also be written in Cartesian coordinates as:

    I = Integral over x, y, z of exp(-x^2 - y^2 - z^2)dx dy dz

    The integrals are all from minus to plus infinity. This integral factorizes:

    I = (I_1)^3


    I_1 = Integral from x = minus to plus infinity of exp(-x^2) dx

    I_1 can be computed by considering it's square and then writing that as a double integral, swithcing to polar coordinates and then computing that integral.

    You find that I_1 = sqrt(pi)


    I = pi^(3/2)

    Then, from the fact that:

    I_1 = Integral from x = minus to plus infinity of exp(-x^2) dx =


    it follows that:

    Integral from x = minus to plus infinity of exp(- p x^2) dx =

    sqrt(pi) p^(-1/2)

    If you differetiate both sides w.r.t. p, you get:

    Integral from x = minus to plus infinity of x^2 exp(- p x^2) dx =

    1/2 sqrt(pi) p^(-3/2)

    This then implies that:

    I = Integral from r = 0 to infinity of A r^2 exp(-r^2) dr

    = A/4 sqrt(pi)

    Equating the two expressions for I yields the equation:

    pi^(3/2) = A/4 sqrt(pi) ---------->

    A = 4 pi

    The volume of the sphere is thus given by 4/3 pi r^3.
  9. Apr 19, 2009 #8
    pls. note, major edit to my post above. I forgot a critical step and have fixed it now.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook