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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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    Hoping not to misread the question, Is the question; a how-to integrate the annuli in the volume torus equation? volume; annuli-filled torus; =(R)(pi^2)(r^2)+(pi)(r^2)(2r) =or=(R)(pi^2)+(pi)(2)(r+r^2) area; annuli-filled torus; =(2)(pi^2)(r)(R)+(pi)(r^2) The radii along the circumference HAS TO BE factored in, otherwise it is just a disk or cylinder shape. I am trying to figure out what a 'annuli-filled' torus is called? It should be a Frisbee shape, but most appear to misinterpret as just a disk shape. It is not just a disk because of the radii along circumference. Such as the difference betw2014-12-09

2 Answers 2

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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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    And you probably know how to evaluate that particular integral. Hint: it is the area under a certain curve, what is the curve?2011-02-10
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Another totally different solution: using a torus-specific change of variable: $\eqalign{ x &=(R+r\cos s)\cos t,\cr y & =(R+r\cos s)\sin t,\cr z & =r\sin s,\cr s,t&\in[0,2\pi]. }$