A solid is generated by rotating about the x-axis the region under the curve $y=f(x)$, where $f$ is a positive function and $x≥0$. The volume generated by the part of the curve from $x=0$ to $x=b$ is $2^b$ for all $b >0$. Find the function $f$.
I understand solids of revolution fairly well and I've managed to set up the following equality:
$$\pi \int_0^b f^2(x)dx = 2^b$$
After some fiddling I am able to discover a function, $f(x) = \sqrt{ln(2)2^x/\pi}$, however when I plug this function back into the formula for the solid of revolution I always get a value which is equal to $2^b -1$ for all b>0.
How do I compensate for the missing one?