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

1. watchmath

We revolve around which axis (line) ?

2. anonymous

Sorry, y-axis

3. watchmath

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

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

Honestly I don't know. You can double check your answer with wolframalpha computation :D.

6. watchmath

the answer should no depends on theta. Maybe you forgot to plug in the limit of integration.

7. anonymous

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

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

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

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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