why is the cone one third of the cylinder? i ask for a new one...?

Question

well, i already had a prevoius answer and i am satisfied with that but i want a new one(more profound and greater). hope you can find one. thnks

Answer 1

Well, you've seen the proof based on the methods of calculus, so let's try a more intuitive (but much less rigorous) approach:

Each horizontal slice of the cone is a circle having a certain area -- consider replacing each slice with a square of the same area. This will change the cone into a square-bottomed pyramid of the same volume. Doing the same with the cylinder, the cylinder is transformed into rectangular cuboid of the same volume, and having the same base and height as the pyramid created from the cone. So to discern the relationship between the volume of a cone and the volume of the cylinder with same base and height, it suffices to find the relationship between a square-bottomed pyramid and rectangular cuboid of the same base and height.

Now, suppose the pyramid has a height h and a base of width w, so the area of the base is w². Now, a rectangular cuboid of the same dimensions can be produced by scaling the rectangular cuboid of width w and height w/2 by a factor of 2h/w, and so unsurprisingly has a volume exactly 2h/w times that of the rectangular cuboid of width w and height w/2. However, the volume of the pyramid of height h and width w has a volume of 2h/w times the pyramid of height w/2 and width w, for exactly the same reason. Therefore, the relationship between a square-bottomed pyramid of height h and width w and the corresponding cuboid is the same as that between the pyramid of height w/2 and width w, and its corresponding cuboid. So all we have to do is consider the special case where the pyramid is exactly half as tall as it is wide.

Finally, consider that in this special case, we may assemble 6 copies of the pyramid into a cube of width w as follows: place one copy at the base, then for the second copy, rotate it 90° vertically, so its tip points parallel to the ground, and then place it on one of the sides of the pyramid at the bottom so one of its triangular faces lies against one of the triangular faces of the pyramid on the bottom, and their tips coincide. Then place the third, fourth, and fifth pyramids on the other triangular faces of the bottom pyramid in the same fashion. Finally, complete the construction by placing an inverted copy of the pyramid on top. If you've visualized this correctly, each face of the cube should be the base of one of the pyramids. This demonstrates that the volume of the pyramid is exactly one sixth the volume of the cube, or w³/6. However, the volume of the cuboid of width w and height w/2 is clearly w²*w/2 or w³/2. We therefore conclude that the volume of the square-bottomed pyramid is exactly one third the volume of the corresponding rectangular cuboid, and that by the argument in the first paragraph, the volume of a cone should be one third the volume of the corresponding cylinder.

As I said, this proof is hardly rigorous, but if you can carry out the visualization, it should be "obvious" why the proportion should be one third and not some other number.

- Pascal

Answer 2

Slice the cone into n pieces with n-1 equally spaced planes parallel to
its base. Each of these pieces is a frustum of a cone, and has volume
that is greater than the volume of a cylinder whose radius is the
radius of the smaller base, and smaller than the volume of a cylinder
whose radius is the radius of the larger base. That is because the
small cylinder is contained in the frustum of the cone contained in the
large cylinder. Here is a diagram of a cross-section of the kth piece,
from the side:

A B C D E
o--o---------------o---------------o--o
| /| | |\ |
|/ | | | \|
o--o---------------o---------------o--o
F G H I J

The cross-section of the frustum is FJDB; of the larger cylinder it is
FJEA; and of the smaller cylinder it is GIDB. A segment of the axis of
the cone and the cylinders is CH, whose length is h/n, the height of
each frustum and each cylinder. For the kth piece from the top, the
radius of the smaller base BC = CD = r*(k-1)/n, and the radius of the
larger base FH = HJ = r*k/n. Then, if V(k) is the volume of the kth
frustum, we have the inequality

Pi*[r*(k-1)/n]^2*(h/n) < V(k) < Pi*[r*k/n]^2*(h/n),
(k-1)^2*Pi*r^2*h/n^3 < V(k) < k^2*Pi*r^2*h/n^3,

Now, add up these inequalities for k = 1, 2, 3, ..., n to get the
volume of the cone in the middle,

V = V(1) + ... + V(n).

Divide by Pi*r^2*h, and multiply by n^3, and you have:

0^2 + ... + (n-1)^2 < V*n^3/(Pi*r^2*h) < 1^2 + ... + n^2.

Now you have to know the sum of the first n squares is equal to
n*(n+1)*(2*n+1)/6, so the sum of the first n-1 squares is equal to
(n-1)*n*(2*n-1)/6. This can be proved by Mathematical Induction if you
don't already know this formula. So,

(n-1)*n*(2*n-1)/6 < V*n^3/(Pi*r^2*h) < n*(n+1)*(2*n+1)/6.

Now, divide by n^3, and rearrange things a little:

(1-1/n)*(1-1/[2*n])/3 < V/(Pi*r^2*h) < (1+1/n)*(1+1/[2*n])/3.

Now recall that n could be any positive integer, and the above
inequalities must be true. As n gets large, the quantity on the left
gets closer and closer to 1/3, and the quantity on the right also gets
closer and closer to 1/3. The difference between these two quantities
is just 1/n. As n gets large, this difference gets small, approaching
zero. In fact, if you want the left side to be larger than 1/3 - a, for
some real number a > 0, just pick n > 1/(2*a), and if you want the
right side to be smaller than 1/3 + a, just pick n > 1/(3*a). That
means that for any positive real number a, we have

1/3 - a < V/(Pi*r^2*h) < 1/3 + a

The only real number x that satisfies 1/3 - a < x < 1/3 + a, for every
positive real number a, is 1/3 itself. Thus

V/(Pi*r^2*h) = 1/3,
V = (1/3)*Pi*r^2*h.

By the way, a very similar proof, depending on the same formula for
the sum of the first n squares, works for pyramids, too.

Sorry the diagram above didn't turn out right. It look fine before I submitted it.

Answer 3

A cone is ⅓ of the cylinder
A pyramid is ⅓ of a block having base area = upper area
Generally, from the central point of the upper area of any regular volume to th sides of the base ,if the volumes re removed actually two third of the volume is removed. This can be proved through calculas.

Answer 4

this is very simple

if a cone and cylinder have the same radius and height, then you will find that the volume of the cone is one third of the cylinder, this can be observed by finding a cone and cylinder with the same height and radius and filling them with rice, you should have one third of the rice in the cone that you have in the cylinder

Answer 5

The volume of a cone, like the volume or area of most objects, can be derived from methods usied in calculus. That doesn't really answer your question, but the formulas is not arbitrary.

Answer 6

God ordained it when the World was created and as long as you are involved with ordinary geometry, you don't have much choice. Just like PI (3.14159.....) is a really inconvenient number, but you can't legislate that the circumference of a circle is 3 times the diameter.

Answer 7

because 3 measurements (radius, radius, and height) are brought together and compressed into 1 point.

like triangle=rectangle/2
because 2 measurements(height/width and base/length) are brought together and compressed into 1 point where height stops and base starts(going down the point).