Can anyone help me with finding the volume of a solid of revolution of f(x) about the x axis for the interval [1,6]. It's supposed to be able to be done without needing calculus but I am having trouble figuring it out.

$f(x) = \begin{cases} 1 & 1 \leq x< 2\\ 1/2 & 2 \leq x< 3\\ . & .\\ . & .\\ 1/n & n\leq x< n+1\\ \end{cases}$

I know the volume would be found like this $\pi$ $\int_{1}^{6}(f(x))^2dx$ but I am unsure about how to go about it with this function.

Any help is appreciated. Thanks

  • $\begingroup$ Well the function is constant in the five intervals $(1,2),(2,3),\cdots,(5,6)$ and the last contribution will for example be $\;\displaystyle\pi\int_{5}^{6}(1/5)^2\,dx$. Further I think like Fantini that $2$ should be $1/2$. $\endgroup$ – Raymond Manzoni Apr 27 '14 at 22:18
  • 1
    $\begingroup$ If that's the case then shouldn't it be $f(x) = 1/2$ for $2 \leq x < 3$? $\endgroup$ – Mark Fantini Apr 27 '14 at 22:19

$f(x)$ is constant in $n$ intervals. Hence, $\int^b_a (f(x))^2dx=(f(x))^2(b-a)$. So, the volume is simply $$\pi\left(1^2(2-1)+2^2(3-2)+\cdots+\dfrac{1}{n^2}(n+1-n)\right)=\pi\left(1^2+2^2+\cdots+\dfrac{1}{n^2}\right)$$ Though I believe that the $2$ should actually be $\dfrac{1}{2}$.

  • $\begingroup$ Thank you I am starting to understand it now. Could you explain how to find the surface area as well on this interval? $\endgroup$ – user2989591 Apr 28 '14 at 0:03

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.