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The question is this

Please tell me if what I did is correct or if there's any faster alternatives.

I set $x$ and $y$ axes on the center of the circle with radius $r$, therefore this can be seen as an area described by $x^2+y^2=r$ revolving around $x=-R-r$

$dV$ can be written as $$dV = 2\pi(R+x)\cdot 2y \cdot dx =2\pi(R+x)\cdot 2 \sqrt{r-x^2} \cdot dx$$

$V$ is then
$$ \int_R^{R+2r} 4\pi(R+x)\sqrt{r-x^2} dx $$

Is this correct?

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You should have $r^2$ under the square root. Your integration range is not correct, it's a range for $x$ and thus should be $[-r,r]$. Anyway, isn't an integral a bit of overkill for this? A bit of thinking can solve the problem with elementary geometrical formulas. – Raskolnikov Feb 9 '11 at 17:36
@Raskolnikov: it is probably a homework question, and since the question explicitly asked for setting up an integral, I think the OP may not get credit if he doesn't do it that way. – Willie Wong Feb 9 '11 at 17:41
Duplicate question:… – TonyK Feb 9 '11 at 18:07
@Raskolnikov: So if use the shell method, the integral should be $$ \int_{-r}^{r} 4\pi(R+x)\sqrt{r^2-x^2} dx $$ ? – xiamx Feb 10 '11 at 15:40
Yes, I think that should be it. – Raskolnikov Feb 10 '11 at 17:08
up vote 1 down vote accepted

hint: in addition to what Raskolnikov said, it may be simpler to consider the washer method instead of the shell method for setting up your integral. (That is, integrate in $y$ and consider the area of annuli of the constant $y$ slices.)

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I tried the washer method and found $$ V= 8\pi R \int_0^{r} \sqrt{r^2-y^2} dy $$ – xiamx Feb 10 '11 at 15:37
And you probably know how to evaluate that particular integral. Hint: it is the area under a certain curve, what is the curve? – Willie Wong Feb 10 '11 at 16:59

Another totally different solution: using a torus-specific change of variable: $$\eqalign{ x &=(R+r\cos s)\cos t,\cr y & =(R+r\cos s)\sin t,\cr z & =r\sin s,\cr s,t&\in[0,2\pi]. }$$

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