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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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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$f(x)$ 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 $\dfrac{1}{2}$.

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  • $\begingroup$ 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

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