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(4/3)(pi)(r^3) = Volume of a Sphere
right?
So why is (4/3) a factor? What is the significence of 4/3? I never really understood that.
I tried to prove that theory of sphere volume, by using areas of 2D cirlces on top of one another. However I did not know the formula for that. So how can you prove the sphere volume theory?
i think you have to use intergration (calculus) to prove it.
Divide a sphere into an infinite amount of pyramid-like solids with a vertex at the centre and infinitely small bases on the surface of the sphere. Let the area of the base of one of these solids be B.
The volume of one of these solids is (1/3)*B*r, where r is the radius of the sphere, and also the height of one of the pyramid-like solids. Since the surface area of a sphere is 4*pi*(r^2), there are a total of (4*pi*(r^2))/B pyramid-like solids. Therefore, the total volume of the sphere is:
(1/3)*B*r*((4*pi*(r^2))/B)=(4/3)*pi*(r^3)
QED:D
knowing that you have some calculus knowledge Kalkewl8ter, if you have the function
f(r)=4pi(r^2)
if you intergrate it:
F(r)=I f(r)dr=(4/3)pi(r^3)
in other words, the derivitive of F(r) is f(r)
That's cool, bugz!
Btw, can anyone explain why integrals are related to area, and derivatives are not, yet they are opposites of eachother?
its is just a geometric interpretation. as long as you know how to do intergrals, you are ok (just like as long as you know that 2+2=4 and not 5, you are ok)
Unless of course so guy is standing over you with a pain generator...
Bugz, you are REALLY bitter at English!:D
Kalk, for your solution thingy. Are the bases of the pyramids with area B rounded? I dont see how to volume is the same if they are and I don't see how your thing makes sense if they aren't. Sorry i'm not as bright.
I'm still confused here. :confused:
I can understand using pyramid volumes in a sphere, but what the hell did you just say?
Give me a better extended Equation with no Calculus or Geometry. And what the hell are Intergals?
Integrals are part of calculus used to do who knows what. Bugz and kalk are just trying to feel special.
So let me get this right.
You have infinit amount of Pyramids with their bases on the outside of the sphere and their points at the center of the sphere.
You stack them in all 4 directions of them selves, their total volume of all pyramids is equal to the sphere's volume.
But still how would it be done? and what is the equation of a half circle; the equation of the parabola mapped on a 2D graph that is equal to a half circle?
Y = Sqrt (Diameter^2 - (X^2)) is the code for maping a half cirlce and doing it negative makes the complet circle. Right? So if you stack areas of circles on the Z Axis that matches the width from side to side of the parabola's x/y postitions. You should also get the area of a sphere.
I think I got it.
How's that for Algebra Kalk? A-l-g-e-b-r-a NOT CALCULUS!
Kalk and me ARE special, and so are you, sprite
Sprite? I'm not a three-dimensional object!Quote:
Originally posted by bugzpodder
Kalk and me ARE special, and so are you, sprite
Algrebra? I used algebra for my entire solution. It's bugz that notices these weird smart little things with calculus. And sprite, the bases don't have to be rounded, because they're so infinitely small that it doesn't matter. (lim B-->0) Just as with the essential idea of integral calculus, a surface with a curved perimeter can be split into many many rectangles with infinitesimally small width.
actually mrs romiens showed me that :D :D :D
Who cares about the Romiens Integral Theory :rolleyes:
just because you don't care doesnt mean other ppl don't care! WAY TO BE AN ASS!
aint talking to you, buddy! talking to silverspriteQuote:
Originally posted by Q_Me
Sprite? I'm not a three-dimensional object!
you dont have to just accept the definition of integralsQuote:
Originally posted by bugzpodder
its is just a geometric interpretation. as long as you know how to do intergrals, you are ok (just like as long as you know that 2+2=4 and not 5, you are ok)
integral is just another way of writing that stupid sumation notation.... I dont remember it now:D
but to get the volume, you write the volume as a sumation formula and you can then convert that to integral
Hey, calm down people! I wonder how mathematicians ever meet in person without massacring eachother.
Mrs Romiens is not as bad as I at first thought anyways, though, bugz, but I still hate the way she does math.:p
I dont like her :mad: makes my blood boil
i didn't ask you ppl if you like mrs romiens or not to me shes just great she taught me calculus and i actually understood it with only grade 9 knowledge so anyone who says she can't teach is obviously wrong i would never buy that capache
I didn't say she can't teach. I think she teaches well, but I believe that she is discriminating against some of us.
oh ur bro said she sucks at teaching and no one could understand her except for him and a few others
I'm not my bro now am I?
no one said you were. i just don't understand why mrs romiens would some ppl's blood boil.
Umm how come you can't understand that she doesnt like everyone she meets. I dunno, she's nice to me. She just thinks i'm stupid.
she gave you an A+ in math that means she thinks you are pretty smart at least in her class. if she thinks you are stupid the highest mark you are going to get is a 70.
Romiens can teach, just not with the winding function. I'm not complaining about marks anymore, just her love of SOHCAHTOA.
i use doublethink: the ability to hold two contradictary thoughts at one time and believe in both of them. I love SOHCAHTOA and I hate SOHCAHTOA.
Doublethink and math do not mix.:p
they just did:D
Kalkewl8ter: Just noticed your smoke and mirrors post. Very cute, but how do you know that the surface area of a sphere is 4*Pi*r2?
Does anybody know of a geometric or algebraic derivation of either formula? The proof using integrals is not difficult if you know calculus, but otherwise it is incomprehensible.
I know that Archimedes is the first to derive these equations, but i have never seen the proof. you may find it if you do a search
the chinese have developed a truely remarkable proof, which can be found here: http://www.staff.hum.ku.dk/dbwagner/...SPHERE-TX.html
Here is Archimedes' work on spheres, notice that they had taken "advantage of some trigonometry and calculus notation to simplify the derivation", which probably was not what you had in mind: http://www.mathpages.com/home/kmath343.htm
You know, I was reading this thread and I thought, Hey, this thread needs yet another reply.
In simple terms, because I am a simpleton, this is the significance of the 4/3 in the formula for volume of a sphere.
First cut the Sphere in half, (with a sharp knife) blah blah, the height is now equal to the radius. So from here on out, when you see an r^3, I have skipped the step of calling the height h and just combined it with the radius r.
Okay, here is the formula for volume of a sphere = 4/3 * pi * r^3
Working backwords, the 4/3 is actually 2 * (2/3 * pi * r^3).
The (2/3 * pi * r^3) is the volume of a hemisphere, which is, as you know, half of the sphere.
So where the hell did the 2/3 come from? That was probably your next question, right?
Well the area of the hemisphere is the volume of a cylinder - the volume of a cone. so....
Volume of a cylinder = pi * r^3 , it is pi * r^2 * h but we are combining it to r^3, since the height is the same as r.
Volume of a cone = 1/3 * pi * r^3.
so... (pi * r^3) - (1/3 * pi * r^3) = 2/3 * pi * r^3, and if you times that by 2, you get 4/3!!!!
Is that the answer you were looking for??
why don't derive what you said! we are more interested in proofs than answers.prove that the volume of the hemisphere of radius r is the difference between the volume of a cylinder of base radius r and the volume of a cone of base radius r!Quote:
Well the area of the hemisphere is the volume of a cylinder - the volume of a cone. so....
Well, I thought the question was:
So, I answered the 4/3 issue.Quote:
So why is (4/3) a factor? What is the significence of 4/3? I never really understood that.
As for why the volume of a hemisphere is equal to the volume(cylinder) minus the volume(cone) question...
I shall close my eyes an hummmmmmmmm, and try to summon Euclid...
Euclid said that any cone is a third part of the cylinder which has the same base with it and equal height. And doubling this gives our sphere. Euclid proved this by proving that a cone could not be more than 1/3, and proved that it could not be less than 1/3.
Not the proofs you were looking for, I know, but who can spit that proof out??? (with out a little refreshing)???
After I take a look at it again, I'll attempt to do the proof. :p
Until then, :)
Archimedes first derived these formulas.
Here is the complete proof, I added the previous proof of the 4/3 figure into it......
Proving the Volume formula for a Sphere, 4/3 * pi * r^3
For the following equations, the heights for the cone and cylinder, h, will be the same measurement of the radius r.
Cone Volume = 1/3 * pi * r^3
Cylinder Volume = pi * r^3
The above are accepted and understood, and will not be proved here.
(Refer to the attached image).
Okay, here is the formula for volume of a sphere = 4/3 * pi * r^3
Working backwards, the 4/3 is actually 2 * (2/3 * pi * r^3).
The (2/3 * pi * r^3) is the volume of a hemisphere, which is, as you know, half of the sphere.
The 2/3 part comes from the area of a hemisphere is volume of a cylinder - the volume of a cone (hemisphere has radius r, which is equal to height of cylinder and cone).
The idea to be proven is that the volume of the hemisphere (one half of the sphere) can be figured by calculating the volume of a proportional cone and proportion cylinder, with the height of each being equal to the radii.
The image is a vertical cross section of the 3d cones, sphere and cylinder. Looking at the picture, the square represents the section of the cylinder, the crossed X represents the two cones, and the circle represents the sphere.
Looking at the three solids, take a horizontal cross-section between C and E. This cross-section intersects the three solids, giving three concentric circles, with the radii of CD (cone), CB (sphere), and CE (cylinder).
The areas of these circles are proportional to the squares of the radii.
Area of a circle = pi * r^2
It is now observed that the squared radius of the cylinder has to be the squared radius of the cone plus the squared radius of the sphere.
Now for connecting the dots ...
(The following are lines; they are represented with the omitted -- on top).
CE = AF = AB
CA = CD, so CA^2 = CD^2
CA^2 + CB^2 = AB^2, and we see above that AB = CE, so AB^2 = CE^2
So therefore, since CA^2 = CD^2
CD^2 + CB^2 = CE^2
So now a proportion is clearly established.
Circle (cone) + circle (sphere) = circle (cylinder), so,
Circle (sphere) = circle (cylinder) - circle (cone).
The Volume of Cylinder (pi * r^3) - Volume of Cone (1/3 * pi * r^3) = 2/3 * pi * r^3, so,
Multiplying by 2 (for two hemispheres) gives 4/3 * pi * r^3.
:p
You know what, I think we need a forum for ppl to discuss how amazingly cool their teachers are, maybe we'll call it the Windsor, Ontario forum! That way, every other thread in this forum won't have hundreds of posts from people arguing about teachers who 90-something percent of the people here dont know!
think you should know them, cuz they are part of the best teachers in the world!
So bugz,
whatchya tink?
i've briefly looked at it and it looks good. but i want to take a closer look tomorrow (with paper and pen and such). but good work!
I made a notation error:
It should read :Quote:
Working backwards, the 4/3 is actually 2 * (2/3 * pi * r^3).
Code:Working backwards, the 4/3 is actually (2 * 2/3) * pi * r^3.
Nice proof, Jared!:D
I tried to develop a similar proof for the surface area of a sphere to the one that I had for the volume of a sphere, and there's something wrong with it, because the result is off. Here it is:
Draw in an infinite amount of "vertical' circumferences equidistant at the equator of the sphere all passing through two points, one at the top and one at the bottom. Then draw in the "horizontal" circumference at the equator of the sphere. Now there is an infinite amount of approximations of isosceles triangles each bounded by two (quarters of) vertical circumferences and (a small section of) the cirumference at the equator. Let the base of each of the triangles be b, where b approaches 0. The height of each of these triangles approaches a quarter of any circumference, which is pi*r/2. The total number of triangles is 2*(2*pi*r/b), and the area of each triangle is (1/2)*b*(pi*r/2). By this logic the surface area of the sphere should be 2*(2*pi*r/b)*(1/2)*b*(pi*r/2)=pi^2*r^2.
Why am I off by a factor of 4/pi?
You lost me again. :confused:Quote:
Originally posted by Jared
Here is the complete proof, I added the previous proof of the 4/3 figure into it......
Now for connecting the dots ...
(The following are lines; they are represented with the omitted -- on top).
CE = AF = AB
CA = CD, so CA^2 = CD^2
CA^2 + CB^2 = AB^2, and we see above that AB = CE, so AB^2 = CE^2
So therefore, since CA^2 = CD^2
CD^2 + CB^2 = CE^2
So now a proportion is clearly established.
Circle (cone) + circle (sphere) = circle (cylinder), so,
Circle (sphere) = circle (cylinder) - circle (cone).
The Volume of Cylinder (pi * r^3) - Volume of Cone (1/3 * pi * r^3) = 2/3 * pi * r^3, so,
Multiplying by 2 (for two hemispheres) gives 4/3 * pi * r^3.
:p
Using the attached image on my earlier post, evaluate each mathematical statement.
Each statement will build upon a previous statement to lead to the conclusion. I broke it down into simple statements, so each one shouldn't be too great of a leap of understanding.
Give me a specific line where you are stuck, and I will explain it.
:)