Welcome to the forum.

Why do I get the wrong answer

If you rotate an isosceles triangle by 2pi you are generating the volume twice over as each half-triangle will make the whole solid.

Split the cross section in half so you have just a right angled triangle, and rotate that.

Bob

]]>integral from 0 to h of 2pi(h/r)x^2 dx = (2pi*h*r^2)/(3). It comes out double.

]]>Welcome to the forum.

The only way I've ever seen this formula derived is with calculus so you have asked something that may be possible.

Do you know the formula for a cylinder?

Now if you had a cone with the same base radius and height, the volume would be less, wouldn't it.

seems reasonable, doesn't I?

I can take you step by step through the calculus if you wish. That way you'll learn some calculus too.

Bob

]]>Can anyone help me, how to derive the formula of the volume of the cone 1/3*pi*r^2*h ?

The volume of a right circular cone with radius r and height h, equals the

area of the right triangle (let the base = r and the height = h), which is

being revolved along the line containing the line segment h, multiplied by the

circumference using the r/3 part of the centroid* as the radius of revolution.

The centroid of a triangle is where all of its medians intersect.

The centroid is the geometric center of the triangle.**

Then the formula for the volume is

the area of the triangle, multiplied by the circumference at the

geometric center (centroid), and using r/3 as the radius of revolution.

This is:

- - . . - - . . - - . . - - . . - - . . - - . . - - . . - -

* Suppose a right triangle is situated on the xy-plane with the

radius extending from (0, 0) to (r, 0) and the height extending

from (0, 0) to (0, h).

The x-coordinate of this centroid is r/3. (This can be worked out using

coordinate geometry.)

** Source:

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

luca-deltodesco has a good proof there. abegale: He has used integral calculus. Maybe you haven't learnt about that yet.

Welcome to the forum.

That formula is just a special case of the formula for any right* pyramid.

Here I am defining a pyramid as any solid where there is a certain shape for the base and then the cross-section as you travel up the axis is a similar shape but reducing at a steady rate until a single point vertex is reached. The shape can be anything; a polygon, a circle, an ellipse or a completely irregular shape. See my picture below.

Suppose the area of the base is S and the height H.

Let s be the area at any point a distance h from the vertex. So S and H are constants determined by the pyramid and s and h are variables.

The ratio of the areas of the small to the large S-shape will be the ratio of the heights squared.

so

So think of a section of the pyramid made by an area s with a thickness of delta h.

Use integration to add up all such volumes thus:

So for any right* pyramid the volume is one third the height times the base area.

*Right here means that the axis is at right angles to the base. It is relatively easy to adapt this proof for non right pyramids.

Bob

]]>Welcome to the forum. What do you not understand?

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line, passing through origin with gradient r/h. rotate graph around the x'axis:

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