Prove that the area under the graph for the function $y=e^{x-e^x}$ is $4\pi^{-1}$ times larger as its volume of revolution

Question

I was meddling around with the chain rule to see if I get somewhere:

Let $A$= area and $V$ = volume

So: I need to prove that: $A=4\pi^{-1}V$ --> $\frac{dA}{dV}= 4\pi^{-1}$

$V=\pi e^{2(x-e^x)}$

$\frac{dA}{dV} = \frac{dA}{dx} * \frac{dx}{dV}$

Therefore:

$\frac{dA}{dV} = \frac{y}{\pi y^2}$ = $\frac{1}{\pi y}$

However I am missing a $4$ here, and I am trying to find out they got it.

Answer 1

\begin{align} A&=\int_{-\infty}^{\infty}\frac{e^x}{e^{e^x}}dx\overset{e^x\to t}{=}\int_{0}^{\infty}\frac1{e^t}dt\\\\ V&=\int_{-\infty}^{\infty}\pi \frac{e^{2x}}{e^{2e^x}}dx\overset{e^x\to t}{=}\int_{0}^{\infty}\pi\frac{t}{e^{2t}}dt\overset{2t\to s}{=}\int_{0}^{\infty}\frac{\pi}4\frac{s}{e^{s}}ds\\ (u=s,v=e^{-s})&=\left[-\frac{\pi}4\frac{s}{e^{s}}\right]_{0}^{\infty}+\int_{0}^{\infty}\frac{\pi}4\frac{1}{e^{s}}ds\\ &=\int_{0}^{\infty}\frac{\pi}4\frac{1}{e^{s}}ds=\frac{\pi}4A \end{align}