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anonymous
 5 years ago
Find the volume of the solid generated by revolving the region bounded by: \[(xh)^2+y^2=r^2,\quad(h>r)\]
anonymous
 5 years ago
Find the volume of the solid generated by revolving the region bounded by: \[(xh)^2+y^2=r^2,\quad(h>r)\]

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1We revolve around which axis (line) ?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Use cyllindrical shell method. The volume is given by the following integral \[2\int_{hr}^{h+r}2\pi x\sqrt{r^2(xh)^2}\,dx\] You can use substitution \(xh=r\sin\theta\) to compute the integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath, 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?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Honestly I don't know. You can double check your answer with wolframalpha computation :D.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1the answer should no depends on theta. Maybe you forgot to plug in the limit of integration.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That 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.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Look 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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you, I guess I missed this part of even/odd functions in the integrand, it's not on my book :(.
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