Question

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

Answer 1

We revolve around which axis (line) ?

Answer 2

Sorry, y-axis

Answer 3

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.

Answer 4

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?

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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
\[4\pi h\int_{-r}^r \sqrt{r^2-u^2}=4\pi h\cdot \text{ area of half circle of radius } r\]
\(=4\pi h\cdot \frac{1}{2}\pi r^2=2\pi^2r^2h\)

Answer 10

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