Question:
Let $n \ge 2$ be a positive whole number. There is a positive function $f_n(x)$ such that when the graph of $f_n(x)$ is revolved around the $x$-axis the volume of the resulting solid between $x=0$ and $x=b$ is $b^n$ for any $b>0$.
$(a)$ Find an expression $f_n(x)$
$(b)$ The region bounded by $f_n(x)$, $y=0$, and $x=b$ is rotated about the $y$-axis. Find the volume of the resulting solid.
My attempt:
(a)
Example... Between $x=0$ and $x=3$, $f_2(x)$ generates a volume of $3^2=9$.
Using the disc method...
Volume $= \int_{lower}^{upper} \pi r^2dx$
$r=$ $f_n(x)$
Volume = $x^n$
$x^n = \int_{0}^{x} \pi (f_n(x))^2dx$
Differentiating both sides...
$nx^{n-1}= \pi f_n(x)^2$ (not sure if my calculus is right here- I do not remember how to differentiate integrals)
Thus,
$f_n(x)= \sqrt{nx^{n-1}/\pi}$
For part b, I don't understand how the region is bound by two different variables ($x$ and $y$)
We know that $b^n = \int_{0}^{b} \pi (f_n(x))^2dx$. Define a function $f: [0, \infty) \to \mathbb R$ by $$f(y)= \int_{0}^{y} \pi (f_n(x))^2dx$$ Then $f(y)=y^n$. Differentiate both sides with respect to $y$ and you will be able to solve for $f_n(y)$ in terms of $y$ to obtain the formula mentioned in the comments.