This is Additional Problem $26$ in Chapter $7$ of Simmons Calculus: "A solid is generated by revolving about the $x$-axis the area bounded by a curve $y=f(x)$, and the lines $x=a$ and $x=b$. Its volume is $\pi(b^3-b^2a)$ for all $b>a$. Find $f(x)$."
Some false starts:
Since $V=\pi b^2(b-a)$ the impulse is to see this expression as the volume of a cylinder with radius $b$, but $f(x)=b$ would yield the correct volume for only intervals $[a,b]$ and not for any other $[a,b']$. So $f(x)$ is not constant and the volume is not a cylinder.
If we say that $g(x)=[f(x)]^2$ and $\int g(x)=G(x)$ then $G(b)-G(a)=b^2(b)-b^2(a)$ and it would seem that $G(x)=b^2x$ but then $g(x)=b^2$ and $f(x)=b$ which is incorrect. So the simple assignments to $G(x)$ aren't right, but I also see no way to disentangle $a$ from $b$.
Rewriting slightly with the disc method we have $\int_{a}^{b} [f(x)]^2dx=b^3\frac{b}{b}-b^3\frac{a}{b}$, implying the volume grows linearly from $0$ to $b^3$ as $a$ goes from $0$ to $b$. In particular, if $a=0$, then $\int_{0}^{b} [f(x)]^2dx=b^3$ suggesting $\int [f(x)]^2dx=x^3$, $[f(x)]^2=3x^2$, and $f(x)=\sqrt{3}x$. But then $\int_{a}^{b} [f(x)]^2dx = b^3-a^3$, wrong again.
I'm overlooking something, or I've made an error, and would appreciate hints, corrections, or solutions.
So we want a function $f(x)$ such that $\displaystyle\int_{a}^{b} [f(x)]^2dx=b^3-b^2a$ for every $b>a$, and $a$ is fixed (I'll assume $a>0$).
Define $F(x)=\displaystyle\int_{a}^{x} [f(t)]^2dt$.
By the Fundamental Theorem of Calculus we will have $F'(x)=f(x)^2$.
But by the above equality we have $F(x)=x^3-x^2a$, and so $F'(x)=3x^2-2ax$.
Thus (assuming we want a continuous function), $$f(x)^2=3x^2-2ax\implies f(x)=\sqrt{3x^2-2ax}\space\text{ or } f(x)=-\sqrt{3x^2-2ax}$$
Both functions are well defined as the only roots of $3x^2-2ax$ are $0$ and $\dfrac{2a}{3}<a$.