# Proving the Volume of a Sphere using Multiple Intergration

1. Feb 7, 2007

### ashnicholls

1. The problem statement, all variables and given/known data

Using the thechnique of multiple intergration prove that the volume of a sphere is 4/3 Pi r^3.

2. Relevant equations

3. The attempt at a solution

Equation for sphere is x^2 + y^2 + z^2 = r^2

Do I solve it using for example z = root(r^2 - x^2 - y^2)

Or is it best do solve it using angle such as y=r sin (Theta)

I hope this makes sence.

Cheers Ash

2. Feb 7, 2007

### dextercioby

You can't prove that. The volume of a sphere is zero.

3. Feb 7, 2007

### ashnicholls

What do you mean because I am using the axis as the center of the sphere.

i was going to work out one octant and then multiply it by 8.

4. Feb 7, 2007

### Benny

In many places a sphere is regarded as being a hollow ball so that might be why dextercioby said the volume is zero.

5. Feb 7, 2007

### dextercioby

The sphere is a surface, its volume is zero. For nonzero radius, it encloses a nonzero volume of a domain in R^{3}, but that's a different enchilada.

http://en.wikipedia.org/wiki/Sphere

6. Feb 7, 2007

### ashnicholls

ye thanks for that. So just for you dextoer, to find the volume of a solid ball!!!!!!!!

So Benny do you mean working it all out using y=r sin (Theta)

So having limits from 0 - Pi/2

Cheers

7. Feb 7, 2007

### HallsofIvy

dextercioby is complete correct- and evil!

If you are going to use xyz- coordinates to find the volume of a ball of radius R, then you should take z going from $-\sqrt{R^2- x^2- y^2}$ to $\sqrt{R^2- x^2- y^2}$. The projection of the ball onto the xy-plane is the disk (NOT "circle"! Don't want dextercioby complaining!) bounded by $x^2+ y^2= R^2$. Let y go from $-\sqrt{R^2- x^2}$ to $\sqrt{R^2- x^2}$. And x, of course, goes from -R to R.

Last edited by a moderator: Feb 7, 2007
8. Feb 7, 2007

### Benny

IMO if this is not an assessed question and you haven't been taught how to use spherical coordinates then don't do this question. If you use xyz coordinates, even if it's 'do-able,' it'll just be a very long exercise in algebraic manipulation, waste of time if you ask me.

If you use spherical coordinates then x = pcos(phi)sin(theta), y = psin(phi)sin(theta) and z = pcos(theta). The angle conventions vary but if you can visualise the situation then the range of theta and phi should be fairly straight forward.

If I recall correctly, the volume integral will be:

$$V = \int\limits_0^{2\pi } {\int\limits_0^\pi {\int\limits_0^r {\rho ^2 \sin \theta d\rho d\theta d\phi } } }$$

You should check that the limits are correct and that you understand how everything comes together, otherwise the above won't be of much help to you.

Note: I don't believe this is giving too much away. If this is an assignment question full marks would not be given for just that much work and if it's just an exercise question then nothing is gained from just having the answer.