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:


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)


$f_n(x)= \sqrt{nx^{n-1}/\pi}$

For part b, I don't understand how the region is bound by two different variables ($x$ and $y$)

  • $\begingroup$ I do not understand your attempt. $\endgroup$
    – Math Lover
    Jul 9 at 19:37
  • $\begingroup$ use $(a)$ to first find the function $\endgroup$
    – Math Lover
    Jul 9 at 19:39
  • $\begingroup$ I am trying to use the disc method but the problem is too abstract for me to understand. $\endgroup$ Jul 9 at 19:46
  • $\begingroup$ I think the function should be $\sqrt{ \frac{n}{2\pi}} \ x^{(n-1)/2}$. Please integrate and check. $\endgroup$
    – Math Lover
    Jul 9 at 19:56
  • $\begingroup$ How did you get that? Thank you $\endgroup$ Jul 9 at 20:04

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.

  • $\begingroup$ I edited my attempt, would you be kind enough to check it? $\endgroup$ Jul 9 at 21:41
  • $\begingroup$ your attempt seems correct, although I would use different variables for the dummy integration variable and the variable that is the upper bound for the integration. This will make things more clear for whoever is reading the solution. I.e. I would write $$y^n= \int_{0}^{y} \pi (f_n(x))^2dx$$ instead of $$x^n= \int_{0}^{x} \pi (f_n(x))^2dx$$ $\endgroup$
    – Hankry
    Jul 9 at 23:18
  • $\begingroup$ Thank you! Now I am having trouble with part b $\endgroup$ Jul 10 at 7:03
  • $\begingroup$ For part b, I don't understand how the region is bound by two different variables (x and y) $\endgroup$ Jul 10 at 15:23
  • $\begingroup$ The region is bounded by three curves in the plane, the first is the graph of the function $f_n$, the second is the horizontal line $y=0$ which is just the $y$-axis, and the third is the vertical line $x=b$. I would draw all three curves for a fixed value of $n$. I.e. take $n=2$ and then draw $f_2$, $y=0$, and $x=b$ and try to identify the region bounded by these curves. Then think about how you can set up an integral to compute the volume of the solid when this region is revolved around the $y$-axis. After you’ve done a specific case it will be easier to generalize to arbitrary $n$, goodluck! $\endgroup$
    – Hankry
    Jul 11 at 1:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.