Finding formula for parabolic cone with constant ratio of volume to surface area of each circular cross section

Question

looking for some assistance with a problem I've been exploring with a friend for kicks. I haven't taken uni level calculus or algebra in a couple of decades so I'm rusty!

I'm trying to express an equation for a parabolic cone (paraboloid?) where the ratio of volume to surface area of the circle through covering every cross section throughout the height of the 'cone' is constant. Presumably there is some parabola equation that creates a line that if spun around to create a parabolic cone shape would give this outcome.

Any takers?

Answer 1

Such a solid of revolution cannot exist. Let the solid be generated by the rotation around $x$-axis of the curve with equation $y=f(x)$. Then its volume, from $x=0$ to $x=X$ is $$ V(X)=\int_0^X\pi f^2(x)\,dx $$ while the area of the cross section at $x=X$ is $A(X)=\pi f^2(X)$. If $A(X)=kV(X)$ for some constant $k$ and any $X\ge0$, then $A'(X)=kV'(X)$, that is: $$ 2\pi f(X)f'(X)=k\pi f^2(X). $$ Hence, either $f(X)=0$ or $f'(X)=(k/2)f(X)$, that is: $f(X)=Ae^{kX/2}$. But that solution doesn't work, because it gives: $$ V(X)={\pi A^2\over k}\big(e^{kX}-1\big) \quad\text{and}\quad A(X)=\pi A^2 e^{kX}. $$

Answer 2

EDIT2:

(After OP comments on 26 Jan 2021)

Let $X=Z = r,Y=y $ symbol $r$ used for cross section radius.Surface area and volume of revolution are found. The ratio is a constant $c/2$, say

$$ \dfrac{\int\pi r^2 dy}{\pi r^2}= c/2 \tag1 $$

Cancel $\pi$ and differentiate with respect to $y$ using Chain Rule

$$ \dfrac{r^2}{2r \frac{dr}{dy}} = c/2$$

$$ \dfrac{dy}{c}=\dfrac{dr}{r} \to \dfrac{dr}{dy}= \dfrac{r}{c} \tag2 $$

Integrate with BC $y=0, r=a\; $

$$ \dfrac{y}{c}=\log \dfrac{r}{a} \tag 3$$

Or

$$ r=a \;e^{y/c} \tag 4 $$

So it is an exponential curve rotated around $Y$ axis.

To verify the obtained result plug (2) into (1) and simplify

$$\dfrac{\text{Volume}}{\text{Cross-Section Area}}=\dfrac{\pi a^2 \int e^{2y/c} dy} {\pi(a^2 e^{y/c})^2} =\dfrac{ e^{2y/c} \cdot c/2} {(e^{y/c})^2}= \dfrac{c}{2} \tag5 $$

which is a constant. So the profile obtained by rotating around y-axis is an exponential curve. It is not a paraboloid of revolution, nor a cone. The ice-cream cup has bottom diameter $2a$ with an upward flare, tallies OK. A rough sketch:

EDIT3:

( EDIT1: First post before 26 Jan 2021)

To me looks that you mean just a surface of revolution not any parabolic cone.

If a surface of revolution between two z- limits is considered, a single area for comparison.. am not able to figure out what it is. So I assumed that you refer to its swept out surface area. With z-axis as symmetry axis, volume/ area ratio is

$$ \dfrac{\pi\int r^{2} dz} {2 \pi\int r \sqrt{1+r^{'2}} dz} =\text{constant}$$

where $r'=\tan \phi,$ the slope. Applying Quotient Rule reduces it to

$$ r\cdot \cos \phi = c$$

which on integration yields as solution the catenoid of revolution.

With initial condition $z=0, \phi=0$,

$$ r= c\;\cosh(z/c)$$

Answer 3

The ratio is a constant $c/2 =k ,\; c= 2k $, say

$$ \dfrac{\int\pi r^2 dy}{\pi r^2}= k \tag {1a} $$

$$ \dfrac{\pi r^2 }{ \dfrac{d (\pi r^2)}{dy}}= k \tag{1b} $$

Cancel $\pi$ and differentiate with respect to $y$ using Chain Rule as above $ (u/v)'=0 \to \left( u/v \to u'/v' \right) $

$$ \dfrac{r^2}{2r \dfrac{dr}{dy}} = k $$

$$ \dfrac{dy}{2k}=\dfrac{dr}{r} \to \dfrac{dr}{dy}= \dfrac{r}{2k} \tag2 $$

Integrate with BC $y=0, r=a\; $

$$ \dfrac{y}{2k}=\log \dfrac{r}{a} \tag 3$$

Or

$$ r= a e^\frac{y}{2k} \tag 4 $$

So it is an exponential curve rotated around $Y$ axis.

To verify the obtained result plug (2) into (1) and simplify

$$\dfrac{\text{Volume}}{\text{Cross-Section Area}}=\dfrac{\pi a^2 \int e^{ y/k} dy} {\pi(a^2 e^{y/(2k)})^2} =\dfrac{ e^{y/k} \cdot k} {(e^{y/(2k)})^2}= k \tag5 $$

which is a constant.