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I have tried setting up multiple systems of equations using many known volumes but I always seem to come up short. My last attempt was a hollow cylinder but that leaves you with three unknowns in only two sim. equations (for V and S.A). Can anyone help?

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mbstackbm
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    If you have three unknowns and only two equations, you can choose the value of one of the unknowns arbitrarily. This problem doesn’t have a unique solution — you have some freedom of choice. – bubba Nov 11 '21 at 03:57
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Brozovic Nov 11 '21 at 03:58
  • @bubba I overlooked that! Will retry it – mbstackbm Nov 11 '21 at 04:12
  • @Brozovic The problem is only giving us a volume and surface area and asking us to find any random solid of revolution that would fit these measurements. I would assume that entails finding dimensions and its axis of rotation. – mbstackbm Nov 11 '21 at 04:14
  • What are the three variables you use when considering a hollow cylinder? A cylinder is described completely by its height and radius. – Taladris Nov 11 '21 at 04:47
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    As least for bounded shape, this is impossible. This violates the isoperimetic inequality for 3-dimenion: $A^3 \ge 36\pi V^2$. – achille hui Nov 11 '21 at 05:15
  • @achillehui Bless you for reassuring me that I didn't just suddenly go retarded. I posted an answer to this problem, received valid criticism that forced me to delete my answer, and then spent over an hour unsuccessfully trying to conjure an example that remedied the oversight in the answer that I posted. – user2661923 Nov 11 '21 at 13:06
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    @Taladris If the cylinder is hollow, you have to consider both the inner and outer radii. This is similar to dimensioning a thick pipe. – user2661923 Nov 11 '21 at 13:11

4 Answers4

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A cylinder of radius $R$ and height $h$ can be obtained by rotating the graph of $f(x)=R$, $0\le x\le h$ about the $x$-axis so it is a solid of revolution.

Its volume is $V=\pi R^2 h$ and its surface area is $S=2\pi Rh$. Then $V/S=R/2=2$ so $R=4$. Therefore, $h=9/2$.

Note: Usually, in Calculus, the surface area does not include the bounding surfaces (obtained by rotating the endpoints of the graph of $f$). If these surfaces should be included, then $S=2\pi Rh+2\pi R^2h=2\pi R(R+h)$. Since $V=72\pi$, we get $h=\frac{72}{R^2}$ so by substituting in $S$, we get the cubic equation

$$R^3-18R+72=0$$

It does not have any positive solution.

Taladris
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There should be many closed curves satisfying the given constraints.

enter image description here If $A$, $C$, $R_a$, and $R_c$ are the area, circumference, distance from axis to area centroid, and distance from axis to circumference centroid, respectively, we have \begin{align} R_a A=36\\ R_c C=18\\ \text{no part of red region cuts the axis} \end{align}

I have tried choosing the red region as a circle as well as a rectangle, they are not possible. Other configurations are left for others. Calculus of variations might help.

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We can use Pappus's centroid theorems for this purpose.

I will take a torus for example, with a centroid at a distance $R$ from the origin axis. Let's say that my torus radius is $r$. So using the 2 theorems, I find that the area $A$ and the volume $V$ are:

$$A = (2\pi R)(2\pi r)$$ $$V = (2\pi R)(\pi r^2)$$

From your hypothesis, $\frac V A = 2$ so $r=4$, and we can find from there that $R=\frac 9{4\pi}$. There might be other ways to find another solution.

EDIT:

Actually $r>R$ so I don't thinks the solution works. It might be that there is no such solid in the way I tried to solve it...

PC1
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  • This is not a correct solution - I will leave it as it provides some hints on how to solve it though. – PC1 Nov 11 '21 at 04:27
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    So does r have to be > R according to the theorem? It is the first time I come across it and the wikipedia is a bit hazy. – mbstackbm Nov 11 '21 at 04:34
  • @mbstackbm the theorem doesn't state anything but the problem is that the torus section width overlaps itself. It's too wide. – PC1 Nov 11 '21 at 04:55
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You can use the basic idea described by @PC1, based on Pappus’ theorems. But use a cone instead of a torus. I think that will work.

bubba
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