Single-Variable Integration Volume Problem. "When lengths change linearly, areas change quadratically"?

Question

Let $C$ be a cone with base any shape (not necessarily a disk or an ellipse) having area $A$, and height $h$ (that is, the distance from the apex to the plane containing the base is $h$).

Note that a cone contains a straight line from the apex to any point on the base perimeter.

Show that the volume of $C$ is $\dfrac{Ah}{3}$.

(Hint: when lengths change linearly, areas change quadratically!)

This problem was posed in the context of using single variable integration to find volumes.

I have experience with using single variable integration to find volumes, but I have never encountered a problem such as this. I attempted to research more information on the hint provided at the end of the problem, but I can't find anything that discusses such a concept.

I would greatly appreciate it if people could please take the time to explain how to solve this problem using the hint at the end. I would also be thankful if people could be gracious enough to also include the reasoning behind each step of their solution, as this will make it easier for me to follow and learn from any answers.

Answer 1

Hint:

Think of integrating using slices along the height. Notice that all the slices will look the same, but will have different sizes, depending on how far along the height you are. Now come up with a formula for the size of the slice at height $x$. Then integrate.

Answer 2

This comes about by considering slices and add them up respecting their inter dependence.

If height is $h$ and radius $r$ but never vary, that is they are constant and independent then volume is

$$ \pi r^2 h $$

Now there is a relation of dependence

$$ \frac{r}{r_{max}} + \frac{z}{h=z_{max}} =1 \tag{1*} $$

$$ V= \pi r^2 dz $$

Plug in 1* and integrate between limits for $ r, z$ as $ (0,r_{rmax}) ,\, (0, h) $ respectively. The result allows recognition of base and height in the formula and how they are involved together. $ A =$ base area $ = \pi r_{max}^2. $

If area is $a b$ (like ellipse $ \pi a b$ ) and each physical dimension doubles then $2*2 = 4 $ is the new area multiple. Similarly volume for ellipsoid ( $\frac43 \pi abc$) is multiplied by 8.

So if the multiple is $m$, area and volume are getting $m^2 , \, m^3$ times multiplied.

It is seen that $2,3 $ are the degree of the algebraic expression.We do not say they are linear or quadratic.

Using theory of indices we add the indices in multiplication.

Similarly if you have a frustum of cone as count the degree to be $1+2=3 $ for volume

$$ \frac{\pi h} {3} ( r^2+h^2+ r h) $$