How Does the Pappus-Guldinus Theorem Apply to Calculating Volume?

[suspenc3]

http://img244.imageshack.us/img244/9114/centroidsdq6.th.jpg

The book only briefly covers this section and there is an example with a sphere, but I don't really know how to get started.

Heres what I tried:

I basically cut the upper region up into three rectangles, two of them having negative area:

triangle 1)$$A=240(100)=24000mm^2$$
$$\bar{y}=250mm$$
Distance traveled by C $$2\pi(250)=1571mm$$
Volume = $$1571mm(24000mm^2=37704000mm^3$$

And then the same thing was done for the other two rectangles of negative area, and then a total volume was found. From this I found a mass which I think is wrong.

Is this the right approach?

 

[hage567]

I don't really know what the Pappus-Guldinus Theorem is or how to apply it, but couldn't you find the volume by using the differences in areas of circles for the each of the sections of different thickness multiplied by the thickness? Just a thought, maybe it might be useful as a check to see if you get a consistent answer.

 

[suspenc3]

Yeah I could try that, I'd still like to know how to apply this theorem though

 

[hage567]

Sorry, I'm not familiar enough with it to be of much use. I'd be more helpful if I could.

 

[denverdoc]

I looked it up; if I understand it is the Vol=planform times the distance the centroid would travel in sweeping out the object.

What I tried is to break into the three rectangles the centroids of each should be easy to figure by symmetry.

I get 2pi(30*100*(80+15)+250*30(110+125)+60*100*(360+30))=
2pi(3000*95+7500*235+6000*390)=
2pi(285,000+1,762,500+2,340,000)=27,567,475.5mm^3=
.0275m^3

 

[suspenc3]

ahhhhh, yeah that makes sense!

how many kilos are in a Mg?(Megagram?)

1000?

 

[denverdoc]

i have never heard of such a creature, my guess would be the same.