In sphere $r \propto \frac{1}{A}$! How is this possible? What's the wrong here?

Question

Surface area $A$ and volume $V$ of a sphere of radius $r$ are
\begin{eqnarray} A=4\pi r^2,\\ V=\frac{4}{3} \pi r^3. \end{eqnarray}

But then
\begin{align} \frac{V}{A} & = \frac{r}{3}\\ \Rightarrow r &= 3 (\frac{V}{A})\\ \Rightarrow r &\propto \frac{V}{A} \end{align}
so, $r \propto \frac{1}{A}$.
But how is it possible? I can realize $r \propto A$. If $r$ increases $A$ will also increase. But what is the wrong in this calculation. Please help me.

Answer 1

You need to understand more accurately what "proportional to" means. If you say that $f\propto g$, this means that $f$ and $g$ are quantities depending on one variable, say $x$, and there is a fixed number $c$ such that $f(x)=cg(x)$ for all values of $x$.

In your statement that $r\propto\frac{1}{A}$, presumably the variable should be $r$. The statement is not justified by the equation $r=\frac{V}{A}$, because if $r$ is a variable then $V$ is also a variable, not a fixed number.

Your statement that "$r\propto A$ [because] if r increases A will also increase" is also wrong - proportionality is not just a matter of "increasing together". In fact, $A=\pi r^2$, and $\pi$ is a fixed number, so you could say $$A\propto r^2\ ,$$ or you could write the equation as $$r=\frac{1}{\sqrt\pi}\sqrt A\ ,$$ so $$r\propto\sqrt A\ .$$ But it is not true that $r$ is a fixed number times $A$.

Answer 2

Last step is not correct, since $V$ depends on $r$.

Answer 3

If you hold $V$ fixed, then the radius $r$ of the sphere varies inversely with the surface area $A$.