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Volumes by Cylindrical Shells (1 Viewer)

vds700

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My tutor told me that you can use this formula:

Delta V = 2pi(Shell radius)(Shell height)delta x or y

unless they ask you to prove it.

But looking at the examples in Cambridge, they go through and do it from first principles each time. This seems to me like the proper way, but is it OK just to use the above formula in exams?
 

tommykins

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vds700 said:
My tutor told me that you can use this formula:

Delta V = 2pi(Shell radius)(Shell height)delta x or y

unless they ask you to prove it.

But looking at the examples in Cambridge, they go through and do it from first principles each time. This seems to me like the proper way, but is it OK just to use the above formula in exams?
rofl that's what i've been taught the whole time, never been asked to prove it either.

PS. if someone can kindly prove it, i'd love you to bits, i can't seem to find my book.
 

independantz

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I've been told you have to prove it each time.
This is the proof i use:

consider a cylindrical shell of very small thickness (delta)r and height,h.
V= v(outer shell)-v(inner shell)
=(pie)(r+(delta)r)^2(h)-(pie)(r)^2(h)
=(pie)(h)(2r(delta)r +((delta)r)^2)
=(pie)(2rh)(delta)r, since (delta)r is v small, ((delta)r)^2 is negligible.

Yeah that's about it.
 

conics2008

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Hi there, your question regarding cylindrical shells...

Your better off knowing how to derive the formula then remember.. just like what the guy above me did.. that method is just crap

Eg pi S 2xy ...

this only applies to some.. if u look at to graphs getting rotated you wont be able to use it..

you better off.. going finding the inner radius and out radius.. and then finding the area times height .. dx or dy.
 

Just.Snaz

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Actually I've done the HSC papers from 1995 - 2007 and in each cylindrical shell question, I used that formula - it agreed with the Maths Association answers I referred to. And then I used it in my exam marked by a hsc marker and I got it perfectly right.
 

melonkitten

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i decided to use the 'shell' method in my 2u exam for the lolz

/haha
 

vds700

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Just.Snaz said:
I suggest you ask a teacher of some authority and credibility.
yeah i will ask my teacher when i go back to school next tuesday
 

cheney31

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i think if the question says "show that the volume of xxx can be expressed as xxx" then u ought to do it from first principle..
but if it is just find the volume of xxx then u can use shortcuts...
thats what i've been doing.. using the formula is so much easier LOL..saves precious time
 

midifile

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conics2008 said:
Hi there, your question regarding cylindrical shells...

Your better off knowing how to derive the formula then remember.. just like what the guy above me did.. that method is just crap

Eg pi S 2xy ...

this only applies to some.. if u look at to graphs getting rotated you wont be able to use it..
.
Yea you shouldnt just use a formula like that because when you rotate around a line like x=-1 it wont work.

But I just write out delta v= 2pi x radius x height x thickness, and go from there

Then you can just work out the radius and height, and sub them in
 

vds700

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i got a shells question from Arnold. Ex 6.2 Q2a)

im fine with the question, u end up with this integral

I tried using integration by parts, and then substuting u = 4 - y, but both times i got 736pi/15. In the solution, they substitute y = 4 - y', which i dont really understand and get 224pi/15. Could someone pplease explain. Thanks
 
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conics2008

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Hey VDS.. Go easy on the question. This is a simple one to start off with..

a) sketch the required graph.
b) the question says bounded by y=4 and x=2..
c) draw those lines and you see a dodgy triangle.. your actually rotating that crap
d) take small strips with change of dy
e) now your inner radius will become y and out outer radius will become y+dy
f) height will be 2-x
g) using 1st principal or what ever u want to call it, it now becomes

d V = pi { (y+dy)^2-y^2}{2-x} ( keep in mind that its dy that means all in term of y

therefore y=4-x^2 = x=root of 4-y

at the end

lim dy > 0 sigma between 4 and 0

pi S 4y-2y root 4-y

integerate normally both sides

you will get 32- 256/15 =224/15 but dont forget to times by pi..

ok.. integration..

S 4y = 2y^2

S 2yroot 4-y ( now u can take out 2 or leave it )

just use normal integarion S( ax+b)^p/q = (ax+b)^p/q +1 / a(p/q+1) on root of 4-y and use integration by parts//

hope u got it..
 

vds700

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conics2008 said:
Hey VDS.. Go easy on the question. This is a simple one to start off with..

a) sketch the required graph.
b) the question says bounded by y=4 and x=2..
c) draw those lines and you see a dodgy triangle.. your actually rotating that crap
d) take small strips with change of dy
e) now your inner radius will become y and out outer radius will become y+dy
f) height will be 2-x
g) using 1st principal or what ever u want to call it, it now becomes

d V = pi { (y+dy)^2-y^2}{2-x} ( keep in mind that its dy that means all in term of y

therefore y=4-x^2 = x=root of 4-y

at the end

lim dy > 0 sigma between 4 and 0

pi S 4y-2y root 4-y

integerate normally both sides

you will get 32- 256/15 =224/15 but dont forget to times by pi..

ok.. integration..

S 4y = 2y^2

S 2yroot 4-y ( now u can take out 2 or leave it )

just use normal integarion S( ax+b)^p/q = (ax+b)^p/q +1 / a(p/q+1) on root of 4-y and use integration by parts//

hope u got it..
Thanks mate, I was only having difficulty with the integration- I just tried again now and I got the right answer.
 

tommykins

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回复: Re: Volumes by Cylindrical Shells


It is Q4a) of the HSC 2002 Paper.
i) is 1.
ii) and iii) is a tad confusing, I don't have the solutions with me so I'm pretty just doing the question.

My answer is 8pi units^3.
 

Poad

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The answer is 8pi (according to this past HSC book that I have).

So you have it right I guess, just because it doesnt actually ask for the volume, so no units^3
 

tommykins

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回复: Re: Volumes by Cylindrical Shells

Thank you
 

vds700

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Can someone plesse exaplin something in the solution to this question from Cambridge. I'm fine getting upto the integral, but I tried expanding and integrating each part, which is long and tedious. However in the solution, they substitute y = 4 - y', which seems much quicker. Can someone please explain how this method works? I don't understand the substitution they used.T hanks
 

conics2008

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vds700 said:
Can someone plesse exaplin something in the solution to this question from Cambridge. I'm fine getting upto the integral, but I tried expanding and integrating each part, which is long and tedious. However in the solution, they substitute y = 4 - y', which seems much quicker. Can someone please explain how this method works? I don't understand the substitution they used.T hanks
hey yeah same,.. im always wonderin wtf is that y ( dash ) used for... i just integrate normally.. good questions =]

yeah u just expand and integrate
 

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