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anonymous

  • 5 years ago

Find the volume of the solid generated by revolving the region bounded by: \[(x-h)^2+y^2=r^2,\quad(h>r)\]

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  1. watchmath
    • 5 years ago
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    We revolve around which axis (line) ?

  2. anonymous
    • 5 years ago
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    Sorry, y-axis

  3. watchmath
    • 5 years ago
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    Use cyllindrical shell method. The volume is given by the following integral \[2\int_{h-r}^{h+r}2\pi x\sqrt{r^2-(x-h)^2}\,dx\] You can use substitution \(x-h=r\sin\theta\) to compute the integral.

  4. anonymous
    • 5 years ago
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    watchmath, thank you for the hint, do you mind helping me out in the end? I end up with: \[4\pi\cos\theta\left[2hr+r^{2}\right]\]is this correct?

  5. watchmath
    • 5 years ago
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    Honestly I don't know. You can double check your answer with wolframalpha computation :D.

  6. watchmath
    • 5 years ago
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    the answer should no depends on theta. Maybe you forgot to plug in the limit of integration.

  7. anonymous
    • 5 years ago
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    That is fine. I tried wolfram and it blew up :(. I guess it doesn't know when there is so many variables. I think my calculus teacher is trying to kill us :D.

  8. watchmath
    • 5 years ago
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    Look at here. It is basically the same calculation: http://www.phengkimving.com/calc_of_one_real_var/12_app_of_the_intgrl/12_04_finding_vol_by_using_cylind_shells.htm

  9. watchmath
    • 5 years ago
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    First substitute \(u=x-h\) to have a nicer integrale \[\int_{-r}^r4\pi(u+h)\sqrt{r^2-u^2}\,du\] Since \(u\sqrt{r^2-u^2}\) is an odd function, then the integral above is reduced into \[4\pi h\int_{-r}^r \sqrt{r^2-u^2}=4\pi h\cdot \text{ area of half circle of radius } r\] \(=4\pi h\cdot \frac{1}{2}\pi r^2=2\pi^2r^2h\)

  10. anonymous
    • 5 years ago
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    Thank you, I guess I missed this part of even/odd functions in the integrand, it's not on my book :(.

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