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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?

Arturo Magidin
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xiamx
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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 09 '11 at 17:36
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    @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 09 '11 at 17:41
  • Duplicate question: http://math.stackexchange.com/questions/11735/how-to-prove-that-a-torus-has-the-same-volume-as-a-cylinder-with-the-height-equa – TonyK Feb 09 '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
  • "revolving around x=−R−r" -> This should be "revolving around x=−R" –  Dec 21 '12 at 21:04
  • From the illustration style, I believe this is out of Stewart's book. (They tried putting it on the Calc I final where I am for two years and finally gave up because most students didn't get the "geometric area" hint.) @xiamx The shell method is a bit easier to set up than the "washer" method for this, but the "washer" integral simplifies more. Either way, you need a technique you don't generally get in first-semester calc, so you use the area formula for a circle instead. The quick method is to use Pappus' Second (Centroid) Theorem, but that isn't covered in most syllabi nowadays... – colormegone Apr 12 '13 at 18:12
  • How to do in cylindrical and spherical (!!!!) coordinates: http://files.eric.ed.gov/fulltext/EJ720058.pdf. – Martín-Blas Pérez Pinilla Dec 09 '14 at 07:35
  • 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 betw –  Dec 09 '14 at 07:11

2 Answers2

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

Martín-Blas Pérez Pinilla
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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.)

Willie Wong
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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
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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? – Willie Wong Feb 10 '11 at 16:59