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

Sorry for any confusion.In summary, there are a couple of ways to prove that the volume of a sphere is equal to 4/3*pi*r^3 using the technique of double integration. One method involves using the disk method or shell method and integrating over the appropriate limits. Another method involves using spherical coordinates and integrating the function 1 over the volume of the sphere. Both methods yield the same result of 4/3*pi*r^3.
  • #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 don't know what original function to start the double integration on?

Help!

Thank you...
 
Physics news on Phys.org
  • #2
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.
 
  • #3
It says i need to use the multiple integration texhnique...
 
  • #4
What's the equation for a sphere? Integrate that function around a circular region of some radius, r.
 
  • #5
I don't 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)...
 
  • #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:
360px-Spherical_Coordinates_(Colatitude,_Longitude).svg.png


Limits:
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 isn't 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:
  • #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

Where

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)

Therefore,

I = pi^(3/2)

Then, from the fact that:

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

sqrt(pi),

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.
 
  • #8
pls. note, major edit to my post above. I forgot a critical step and have fixed it now.
 

1. How does double integration prove the volume of a sphere?

Double integration is a mathematical technique used to calculate the volume of a three-dimensional object by integrating over two variables. In the case of a sphere, double integration involves integrating over the surface area of the sphere to find the volume enclosed within it.

2. What is the formula for calculating the volume of a sphere using double integration?

The formula for calculating the volume of a sphere using double integration is V = ∫∫dS, where V is the volume, ∫∫ represents the double integral, and dS is the differential surface area element of the sphere.

3. Why is double integration necessary to prove the volume of a sphere?

Double integration is necessary because the volume of a sphere cannot be calculated using a single integral. This is because the volume of a sphere is not a simple function of one variable, but rather a function of two variables - the radius and the angle of rotation.

4. What are the steps involved in proving the volume of a sphere using double integration?

The steps involved in proving the volume of a sphere using double integration are: 1) Setting up the integral by defining the limits of integration for both variables, 2) Evaluating the integral, and 3) Simplifying the result to get the final formula for the volume of the sphere.

5. Are there any other methods for proving the volume of a sphere?

Yes, there are other methods for proving the volume of a sphere, such as using the formula V = (4/3)πr³, where r is the radius of the sphere. This formula is based on the fact that the volume of a sphere is four-thirds of the volume of a cylinder with the same radius and height.

Suggested for: How do i prove the volume of a sphere by double integration?

Replies
14
Views
895
Replies
30
Views
273
Replies
9
Views
719
Replies
23
Views
1K
Replies
8
Views
496
Back
Top