Where am I wrong at deriving the formula of volume of cone?

Question

I was deriving the formula of cone volume yesterday but I was stuck at a place.My reason for asking help in Math SE is because I was not doing it by looking at any book or the internet.So,I want your help to know where my thinking went wrong.

My attempt:

In a right circular cone with base radius $r$ and height $h$,both keep changing as we move upward.So,my integral must take into account the change of height (from $0$ to complete length $h$) and change of radius (from base radius $r$ to $0$ at the apex.)

So $$\begin{align} V_{cone} & =\int_r^0\int_0^h\pi r^2 dr \space dh \\ & =\int_r^0\pi r^2 dr \cdot \int_0^h\pi r^2 dh \end{align}$$

In the first integral I'm treating $r$ as a variable but in the second integral I'm treating it as a constant.

Solving this does not yield the formula $V_{cone}=\frac13\pi r^2h$.

Where I made a mistake?

Answer 1

What you're calculating has unit of $\text{length}^4$ so it can't be volume. Where you went wrong is : $h$ and $r$ are related via $r=h\tan\alpha$ where $\alpha=$ half the angle at vertex. If the base radius is $R$ and height of the base from vertex is $H,$

the volume is

$$ \int_0^H\pi r^2dh=\int_{0}^H\pi\tan^2\alpha \,h^2dh={1\over3}\pi\tan^2\alpha H^3=\frac13\pi {R^2\over H^2}H^3={1\over3}\pi R^2H $$ as $R=H\tan\alpha$.

If you want direct integration use triple integral $$ \int_0^{2\pi}\int_0^R\int_{r\cot\alpha}^H rdh\,dr\,d\theta={\pi\over3}R^2H $$

Answer 2

The better way to cast this problem is in terms of an integral of one variable. Let $H, R$ be the constant height and base radius of your cone, respectively. The radii of the circular cross sections that you make as you move up the $y$ axis of your cone vary linearly between $R$ and $0$. Convince yourself that for a $y$ where $0\leq y \leq H$ that the radius of a cross section $R_c$ at $y$ is $$R_c(y) = \frac{R(H-y)}{H}$$ because when $y = 0$ (bottom of the cone) we have radius $R$, and at $y = H$ (top of the cone) then we get radius $0$. This means the area of such a cross section is simply $\pi R_c(y)^2$. Simply integrate $$\int_0^H \pi R_c(y)^2 \text{d}y$$ to get your result.

Answer 3

There's two things you are doing wrong I think.

In choosing a double integral and integrating twice. Let's think about what you are calculating. The first integral you are integrating the areas of a bunch of circles as r goes from 0 to h. This will be the volume of a "stack" of circles. In essence, it is the volume of a cone where the height is equal to the radius expressed in terms of h. The second integral seems to be taking a constant $r$ (which coincidental is rewritten the same as the variable in the first integral but which can no longer refer to the first variable as the first variable has been "integrated out") and calculating from 0 to r the hyper-volume of a set of four dimensionally packed cones cones varying in height from 0 to r.

The second error is in evaluating the double integral you seem to have just repeated what you are integrating, $\pi r^2$ twice. I ... don't understand why you did that, so I can't really comment on that.

To properly integrate we need to set it up properly. We want to integrate the areas of circles in terms of h as h goes from 0 to $H$ ($H$ = the final Height). i.e. $\int_0^H A_h dh$ where $A_h$ is the area of circle at height $h$.

At height = $h$, where we are $h/H$ of the way up the cone, the radius of that circle will be $r = (h/H)*R$ where $R$ is the final radius.

Now we could do a second integral to determine $A_h = \int_0^{hR/H} 2\pi r \space dr$, that is, the area of a circle is the infinite sum of the circumferences but... c'mon we know the area of the circle; it is $\pi r^2$ and as $r = hR/H$ we have:

Area of Cone = $\int_0^H \pi (hR/H)^2 dh = \pi (R^2/H^2)\int_0^H h^2\space dh$

Which if we did want to do a double integral would be the same as:

Area of Cone = $\int_0^H \int_0^{hR/H} 2\pi r\space dr\space dh$.

If you work them out you should get the area of a cone.