Find a solid of revolution whose volume is 72π and whose surface area is 36π.

Question

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?

Answer 1

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...

Answer 2

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.

Answer 3

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.

Answer 4

There should be many closed curves satisfying the given constraints.

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.