What is the volume of a sphere?

[Jameson]

I'm trying to prove the volume of a sphere is (4/3)(pi)r^3. (Sorry I haven't figured out the tex thing yet)

I was thinking that the volume of a sphere is the sum of the circular cross-sections that make it up. Since "r" is different for each cross-section, you put in the variable "x" and get:

(pi)x^2.

The height of the sphere can be represented by the change in y, (dy) so now we get the integral:

int{ (pi)x^2*dy }

Since we need the variable of integration in terms of y, I went to the equation of a circle.

x^2 + y^2 = r^2

So, x^2 = r^2 - y^2

Substituting that into the integral we get the final intergral:

int { (pi)(r^2-y^2)*dy }

This is as far as I can think this out. Where did I make a mistake or where do I need to go from here?


-------------
Jameson

 

[dextercioby]

Do you know triple integration and change of variables in triple/multiple integrations...?

Daniel.

 

[Jameson]

No I do not. Is that the only way to prove this?

 

[dextercioby]

BTW,in case you didn't know,the volume of a sphere is ZERO...

Daniel.

 

[Jameson]

Look. If you want to prove how smart you are, do it somewhere else. If you just don't know how to do this problem, then don't say anything. If you do, I would like your help.

How is the volume of a sphere zero?

 

[Crosson]

Your method is suitable, buth have you ever heard of spherical polar coordinates?

It is a radial coordinate r, and two polar angles theta and phi. Notice how this coordinate is a much more natural one, for say the motion of you elbow and fore arm. There is some fun geometry to figure out, but here is an appropriate transformation :

x = r sin ( phi) cos ( theta)

y = r sin ( phi )sin (theta)

z = r cos ( phi )

using this, you can transform dx, dy, and dz into dr etc

Then just do a triple integral where r goes from 0 to R, the radius of the sphere. Theta goes from 0 to 2pi and Phi goes from zero to pi. My explanation is not so great, but you should look up spherical polar coordinates.

 

[dextercioby]

I'm saying the volume of a sphere is ZERO,USING THE DEFINITION OF A SPHERE.

As for your initial method,it was incorrect.You were insuccessfully attempting to find the volume of a ball of radius R using cylindrical coordinates...

Daniel.

 

[dextercioby]

http://mathworld.wolfram.com/Sphere.html

You were trying to derive the result using formula #15 and Crosson referred to formula #16.

Daniel.

 

[cepheid]

Look. If you want to prove how smart you are, do it somewhere else. If you just don't know how to do this problem, then don't say anything. If you do, I would like your help.

How is the volume of a sphere zero?

Dextercioby always tells people that what they have said is wrong, based on semantics. That makes it impossible for them to understand why. All it does it confuse the issue. It's not a very good way to confront people who are trying to learn IMHO. Dexter is pointing out that a sphere is a two-dimensional surface in 3D space, therefore it has no volume. However, the volume **ENCLOSED** by that spherical surface is given by $\frac{4}{3} \pi r^3$. Arrrgggh! See what I mean? Frustrating guy! The volume of a spherical solid is given by that formula too.

You CAN use single variable calculus to determine that this formula is correct, i.e. you don't have to perform a triple integration, only a single one. Say you have a sphere centred at the origin. One of those thin (let's make them vertical) slices you were talking about has area pi r^2 (presumably you already derived the formula for the area of a circle. If not, you can use calculus for that too). The slice has an infinitesimal thickness dx. And as you move along the x-axis from -r to r, through the sphere, you're encountering larger and larger slices, with the largest being at the centre, and then smaller and smaller, till you get to the other end. You want to integrate all of these "infinitesimal" volumes. But the radius of each slice is not constant! It is a variable. If you are a distance x away from the centre of the sphere, the radius of your slice is 'y', where y is the y coordinate of the point on the sphere at that x coordinate. Using pythagoras, $y = \sqrt{r^2 - x^2}$

Integrate:

$$V = \int_{-r}^{r} {\pi (r^2 - x^2)dx} = 2 \pi \int_0^r {(r^2 - x^2)dx}$$

EDIT: LOL...i didn't need to explain any of this to you. You had already derived it in your first post! So, have you encountered some specific problem in trying to integrate that?

 

[Jameson]

Thank you very much for your explanation. I don't know how to integrate $$2 \pi \int_0^r {(r^2 - x^2)dx}$$ since it has two different variables, but I can find out how to do that.

Many thanks,
Jameson

 

[cepheid]

r is not a variable. It is the radius of your sphere, hence it is constant. I hope that helps you out!

 

[dextercioby]

It's not SEMANTICS,it's plain (simple) mathematics (which 'encloses' a great deal of logics)...The volume in question EXISTS,but it's zero...

Daniel.

 

[Curious3141]

I'm trying to prove the volume of a sphere is (4/3)(pi)r^3. (Sorry I haven't figured out the tex thing yet)

I was thinking that the volume of a sphere is the sum of the circular cross-sections that make it up. Since "r" is different for each cross-section, you put in the variable "x" and get:

(pi)x^2.

The height of the sphere can be represented by the change in y, (dy) so now we get the integral:

int{ (pi)x^2*dy }

Since we need the variable of integration in terms of y, I went to the equation of a circle.

x^2 + y^2 = r^2

So, x^2 = r^2 - y^2

Substituting that into the integral we get the final intergral:

int { (pi)(r^2-y^2)*dy }

This is as far as I can think this out. Where did I make a mistake or where do I need to go from here?


-------------
Jameson

Jameson, calculus is the "easy" way to do it (that is, if you're an advanced student. ) but I'm guessing you'll love this non-calculus method to the same end :

http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html

This method was originally due to Archimedes. Cool guy.

 

[Gokul43201]

As for your initial method,it was incorrect.

I don't see how. It looks perfectly good to me.

 

[saltydog]

Me neither

Yea, I know I'm slow. Anyway, here's the volume using a triple integral. And I didn't know either what Daniel meant about the volume being zero, and in fact it took me a while to figure it out even after Cepheid explained it.

In spherical coordinates, the problem can be defined as follows:

$$vol=8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\int_0^r \rho^2 \sin(\phi)d\rho d\theta d\phi$$

Beautiful isn't it!

So:

$$8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2} \sin(\phi)(\frac{\rho^3}{3}){|}_0^r d\theta d\phi$$

and then:

$$\frac{4r^3 \pi}{3}\int_0^\frac{\pi}{2}sin(\phi)d\phi$$

or:

$$-\frac{4r^3\pi}{3}[0-1]=\frac{4\pi r^3}{3}$$

Don't you just love Calculus!

 

[Jameson]

Thanks to everyone for their help. I still don't quite understand how "r" is a constant in the integral... that would mean you could pull it out in front of the equation and get:

2(pi)r^2 * int { -y^2 dy }

Is that the correct answer?

Thanks again to all,
Jameson

 

[dextercioby]

No.You'll have to integrate "r^{2}" like any other constant.So if you pull it out of the integral,u must multiply with the length on the interval:

$$C\int_{a}^{b} dy=C(b-a)$$

Daniel.

 

[Galileo]

The radius of the sphere is a given for the problem. It does not depend on x or y.

$$2 \pi \int_0^r {(r^2 - x^2)dx}=2\pi\left(\int_0^r r^2dx-\int_0^rx^2dx\right)$$
Then you can take the $r^2$ out of the first integral.

 

[tongos]

i think one could prove it without calculus.

 

[mathwonk]

jameson, after reading your impatient outbursts at people for rudeness, i do not feel like helping you. when you post on a public forum like this, anyone at all may respond. in my opinion, you are asking for something for nothing, and should be grateful for whatever you get of value.

 

[dextercioby]

i think one could prove it without calculus.

One could use the theorem of Pappus & Guldin,but that theorem woul still require calculus for proving.

Would you care so show us how u thought of it...?

Daniel.

 

[mathwonk]

well it depends what you call calculus. certainly archimedes knew how to do this problem, using only limits. i.e. he approximated it by volumes of cylinders, then took the limit of those approximating volumes, using the formula he knew for the sum of squares of the first n integers: 1^2 + 2^2 + 3^2 +...+n^2 = (n^3)/6 + lower degree terms.


if you consider calculus to denote the relation between differentiation and integration, then this method is not strictly calculus. It is equivalent to summing the "Riemann sum" for the volume integra for the sphere, but directly by taking the limit, without the FTC.

 

[dextercioby]

I think of calculus to start from the concept of sequence of numbers..."Limit" is the essential concept,so yes,even summing a Riemann sum without FTC is still calculus...

That's my view.It's not necessarily the unique/correct one.

Daniel.

 

[saltydog]

well it depends what you call calculus. certainly archimedes knew how to do this problem, using only limits. i.e. he approximated it by volumes of cylinders, then took the limit of those approximating volumes, using the formula he knew for the sum of squares of the first n integers: 1^2 + 2^2 + 3^2 +...+n^2 = (n^3)/6 + lower degree terms.

Did Archimedes actually come up with the exact value or only an approximate one? I'll search the net to see if I can find out (and learn about him too!)because if he only came up with an approximation then that's not solving the problem in my opinion.

 

[dextercioby]

What do you understand by "approximation" in this specific case...?

Daniel.

 

[Jameson]

Mathwonk...

I have made no impatient outbursts. The only comment I made was to a post which did nothing but make an insult towards something I had posted. I am only asking for help. I do not see where you are coming from and I have thanked all who have helped me.

 

[saltydog]

What do you understand by "approximation" in this specific case...?

Daniel.

Alright, it's exact:

"On his grave their is an inscription of pi, his most famous discovery. They also placed on his tombstone the figure of a sphere inscribed inside a cylinder and the 2:3 ratio of the volumes between them, the solution to the problem he considered his greatest achievement."

I'll spend some time learning about him . . .

 

[tongos]

i don't think it can be solved without the use of calculus, as dexter said, once we incorporate limits then the problem does become calculus . Somebody could have no knowledge for calculus, but could still find the volume of a sphere (as i did when i was a freshmen in hs, now I am a senior). Only if that somebody would realize that a limit problem is taking place, in this, the somebody would be doing calculus without knowing it.

 

[mathwonk]

tongos is defining calculus to mean limits. in that case archimedes is responsible for the calculus. I myself tend to disagree with this, and am backed up by some of my friends in math education, who have studied the history of mathematics, but it is really a matter of terminology.

anyway, i could be wrong , but it is my opinion that it is extrnmely well known that archimedes did discover the exact formulas for the volume of a sphere,and surface area of a sphere, (or area of the surface of a sphere of we have any word police out there), as well as the area under a parabola.

he even is said to have directed that his tomb be decorated by a picture illustrating a sphere inscribed in a cylinder as he knew that ratio of their volumes (2/3?).

 

[dextercioby]

Okay,Mathwonk,it is the volume of a ball and the area of a sphere (the 2-sphere is a surface by definition,so it wouldn't make sense to assert:surface of a sphere.It would be pleonastic).
And yes,words 2 matter in mathematics...

Daniel.

 

[cepheid]

Thanks for the interesting link to Archimedes' method. I remember my first year calculus text referenceing the "area" problem as a motivation for integral calculus, but one that had been solved by the Greeks, using the notion of a limit. However, they didn't explicitly define that notion or coin the term, IIRC. It's very interesting stuff. My apologies for my poor explanation regarding the "r" in the integral. It is a variable after all...the method works for any solid sphere of radius r. What I meant was that since we were integrating with respect to x (or y), of which r was independent, it could be treated as a constant in terms of the actual integration. The variable that you integrate with respect to does not remain after integration. For any variables that do remain, the volume is expressed as a function of those variables. So you get a general formula of the volume of a spherical object as a function of radius: V(r). Plugging in a specific value of r gives you a specific volume for a sphere of that radius, as you already know. I hope that clear's things up Jameson. Apologies to Daniel for previous remarks. All I was trying to say was that you were pointing out to Jameson that his use of the term "sphere" was semantically incorrect (if that is the right word...I mean to say that it was the wrong terminology *strictly speaking*), but you didn't state that explicitly, only implied it. Given that it was pretty clear what he meant, ie. that he was trying to find the volume of a spherical solid, or "ball", or the volume enclosed by a spherical surface, or whatever, I thought that not explaining clearly what the error was would only lead to confusion. I'll be honest: I used to think of a cylinder as a can...with a circular base. The mathematical definition is far more broad, as we learned in 3D coordiniate geometry. Same thing with sphere---until we saw the formal definition, I equated the term "sphere" in my mind with the everyday use of the word.

 

[mathwonk]

well to equate limits with calculus sems a bit broad to me, as that poses the question whether it is calculus to asset e.g. that .333... = 1/3, or even if .9999... = 1.

most inclusively, calculus involves three parts, the limit calculation of area, and of tangency, and of the connection between the two.

basically what archimedes did was discover a formula for the sum of the first n squares. then if you write out the Riemann sum - integral formulas for the area under the parabola y = x^2, ot for the volume of a sphere (or ball, but I think the historical justification for this use is lacking), you will see this formula occurs in both. so all three problems are the same.

 

[mathwonk]

assume your sphere has radius R, and you subdivide it into n equal parts each of length R/n, and circumscribe a stack of cylinders each of height R/n, inside the sphere.

Then by pythagoras, the radius r of the ith cylinder satisfies r^2 + [(i-1)R/n]^2 = R^2.

i.e. r^2 = R^2 - [(i-1)R/n]^2.

So the volume of the ith cylinder is pi r^2 (R/n) = pi (R/n) [R^2 - [(i-1)R/n]^2].

Hence the sum of all their volumes is the sum of these volumes as i goes from 1 to n.

which is pi R^3 - pi R^3 [1^2 + 2^2 +...+(n-1)^2]/n^3

now if we have the formula for the sum of these squares, namely (n^3)/3 +
terms of lower degree,

we see that as n goes to infinity, this quantity approaches pi R^3 - (1/3)(pi)R^3

= (2/3)pi R^3 = volume of a hemisphere (or semi - volume enclosed by a sphere).

thus indeed knowing the sum of the squares implies one can compute volume of a spherical solid.

 

[mathwonk]

as to area under a parabola like y = x^2, one subdivides the x-axis between 0 and b, into n equal parts, each of length b/n, and circum scribes rectangles over the parabola in each subinterval, the ith one having area (ib/n)^2(b/n) = i^2 b^3/n^3.

adding up from i =1 to i=n, gives us b^3 [n^3/3 +...]/n^3, which as n goes to infinity, approaches b^3/3.

again the only ingredient of the problem is the sum of the first n squares.

so archimedes made a very intelligent use out of that one formula, plus his understanding of limits.

which pretty much makes him father of integral calculus, if you wish.

i do not understand sow ell how he compouted the surface area of a sphere, unless he did it the way we do in calculus, approximating it by cones. i guess that's it. one of his books was on cones i think. so he was sort of the master of quadratic mathematics.

 

[tongos]

this has nothing to do with limits, but i thought up this way of finding the sum of the squares, which i thought was pretty cool.
1^2+2^2+3^2+4^2+5^2...x^2

all you have to do is look at increasing three digit numbers, and the amount of combinations in an increasing three digit number is 1+2+3+4+5+6+7+(1+2+3+4+5+6)+(1+2+3+4+5)...

but there is any easier way to express this, it could be 9(8)(7)/6 because an increasing number is only one out of six.

now look 100(1)+/99(2)/+98(3)+/97(4)/+96(5)+/95(6)/...+/1(100)/= 102(101)(100)/6

and 99(1)+97(2)+95(3)+93(5)...+1(50)

so the terms in / / are going to be repeated again, and you divide that by two so you get the series below, and you get (102)(101)(100)/24 as the sum which is (2n+2)(2n+1)(2n)/24 which is equal to the sum of the nth squares.
kind of a cool way. to find the sum of cubes, use four digit numbers.