Torus (1 Viewer)

[nerdd]

hey guys..
just wondering how you guys do the torus questions.
by slicing or cylindrical?

the arnold book uses cylindrical and its a bit confusing I think.. especially cause they leave out a lot of lines. (well it is to me ok? )

and then i found these other notes that do it by the annulus way.. and end up with (x1 - x2)(x1 + x2)
but it seems to only work if the circle that you're rotating, doesn't have the centre at (0,0).
because if it is at the origin, (x1 - x2) ends up being zero... so the
pi(x1-x2)(x1+x2) method doesn't work.

any suggestions?

well i tried doing the annulus way to answer this...

'the circle x^2 + y^2 = 25 is rotated around the line x = 10. find the volume of the torus found'

answer is 1000 pi ^2 or 500 pi ^2 sorry i'm having a mental blank and can't remember it exactly.. will edit post when i go and check.

 

[wogboy]

Cylindric shells looks like a good way to find the volume of a torus:

the circle x^2 + y^2 = 25 is rotated around the line x = 10. find the volume of the torus found

Draw up a nice big diagram, with a circle of radius 5 & origin centre. Then draw it's symmetric reflection having a centre at (20,0) and radius 5. Draw in the cylindric shells, with the cylinders oriented "vertically".

dV = 2*pi*r*h*dx

(find the relationship between the cylinder radius/height, and its x position)
r = 10 - x
h = 2*sqrt(25 - x^2), so

dV = 2*pi*(10 - x)*2*sqrt(25 - x^2) dx
V = 4*pi*I{-5 -> 5}(10 - x)*sqrt(25 - x^2)

V = 4*pi* [I{-5 -> 5} 10*sqrt(25-x^2) - I{-5 -> 5} x*sqrt(25-x^2)]

This is where you have to be clever, f(x)=x*sqrt(25-x^2) is an odd function, so I{-5 -> 5} x*sqrt(25-x^2) = 0 by symmetry. Also I{-5 -> 5} sqrt(25-x^2) = 25*pi/2 (area of a semicircle)

so V = 4*pi*10*25*pi/2
V = 500*pi^2 units^3

 

[Affinity]

shells save writing. I prefer not to use the slicing method if I can (for this one)

By the way, for some volumes, heres a way to check your answer:
(you cannot assume this in your answer)

Pappus' theorem:

the volume of any solid of revolution equals

A*(2Pi*r)

where A = area of the region rotated and r is the distance from the centroid of the area to the axis of rotation.

so for this torus,

A= Pi*5^2 = 25 Pi
r = 10
V=25Pi*(2*Pi*10) = 500 Pi^2 unit^3

 

[ezzy85]

are there any programs that can draw these 3d shapes with the shells? im having trouble visualising the shells in some of the shapes.

 

[McLake]

Which shapes? You really need to be able to visulaise "on-the-spot" so that you can answer questions in an exam ...

 

[ezzy85]

how a shell or a washer would look inside a torus. i just need to see a picture of it 3d and that will proably help me for future questions.