Cylindrical shells to find volume of a torus

[robertmatthew]

Homework Statement

Use cylindrical shells to find the volume of a torus with radii r and R.

Homework Equations

V= ∫[a,b] 2πxf(x)dx
y= sqrt(r² - (x-R)²)

The Attempt at a Solution

V= ∫ [R, R+r] 2πx sqrt(r² - x² - 2xR + R²) dx

I feel like this isn't going in the right direction, though. I found a different post about the same question, but that person was using disks, and the volume was 2π²r²R, and I can't see where the π² would come from in my equation, so I think I'm going about this incorrectly. Any help is much appreciated, thanks.

 

[Simon Bridge]

It looks like your integrand only includes the top half of the shell.

consider: the shell at radius x: R-r<x<R+r, is dx thick 2πx around and h(x) tall.
... so it's volume is dV= 2πxh(x)dx

Sketch it in - label h, then label y, and see how they are related.

Apart from that - keep going. You'll probably need a trig substitution.
Maybe x-R = r.sinθ ?

 

[robertmatthew]

Would h be 2y, since the height of the cylinder depends on the curve of the torus? So for any point on the torus, where the circular cross section lies in the xy plane, y=sqrt(r²-(x-R)²). Then the cylindrical shell would dip above and below the x axis, so the height should be 2y, right?

Like this, if that helps clarify at all.

 

[SammyS]

Would h be 2y, since the height of the cylinder depends on the curve of the torus? So for any point on the torus, where the circular cross section lies in the xy plane, y=sqrt(r²-(x-R)²). Then the cylindrical shell would dip above and below the x axis, so the height should be 2y, right?

Like this, if that helps clarify at all.

That looks good to me. It appears to agree with Simon's comment.

 

[Simon Bridge]

That's what I was thinking - nice pics BTW.
There is no arguing with a decent sketch - people used to say "God is a geometer".

So now you have the volume of the cylindrical shell radius x: R-r<x<R+r, is $$dV = 4\pi x \sqrt{r^2-(x-R)^2}\; dx = f(x)\; dx$$... now it is just a matter of adding up all the dV's.
Since these are infinitesimal thickness shells, the summation sign is replaced by an integral over appropriate llmits. $$V=\int_V \; dV = \int_{a}^{b} f(x)\; dx$$ ... writing that out, with correct numbers for a, b, and f(x), is the next step - then you can use a substitution.

 

[robertmatthew]

Alright, I substituted u=x-R and solved far enough to find a shortcut using the area of a semicircle, and then evaluated to get 2π²Rr². Thanks!

 

[Simon Bridge]

I'd have gone right to x-R=r.sin(u).
But well done anyway.

Notice that $A=\pi r^2$ is the area of the crossection circle and $C=2\pi R$ is the circumference at the center. This means the volume of the torus turned out to be V=AC ... which is the volume of a cylinder radius r and height C. Does that make sense?