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Volumes - Slicing (1 Viewer)

haboozin

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hey,

I came along this question. Did it... looked at the solution and they do it differently.
I dont know why they do that and they get a different answer.

So i'll see what you guys get..
I'm pretty sure mine is right
 

KFunk

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Did you get anything in the realms of 2547a<sup>3</sup>&pi;/20 units cubed? I banged that up pretty quickly so chances are it's wrong but if not let me know and I'll type up any part of the solution that you're after.
 

serge

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did anyone get an answer with 3 different powers of a?
 
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haboozin

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KFunk said:
Did you get anything in the realms of 2547a<sup>3</sup>&pi;/20 units cubed? I banged that up pretty quickly so chances are it's wrong but if not let me know and I'll type up any part of the solution that you're after.

my answer is (around) ... half of yours

the answer they gave... much smaller.

i dont think they're right


This is what they do:

(R - r)^2

and it really bothers me, it should be R^2 - r^2
 
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serge

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do they have any working? what book's it from
 

serge

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it's slicing why would your method be correct?

it looks like you're trying to rotate it around something...

post the answer they give
 
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haboozin

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serge said:
it's slicing why would your method be correct?

it looks like you're trying to rotate it around something...

slicing is a method of finding a volume...


a volume is achieved by the shaded area being rotated?

stop confusing me... stressful times
 

serge

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haboozin said:
slicing is a method of finding a volume...


a volume is achieved by the shaded area being rotated?

stop confusing me... stressful times
yeh, my bad

i mean it is a plane question,
I didnt know you could get those by rotating
 
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haboozin

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robbo_145 said:
i get (9&pi;a2)/4

not sure if im reading the question right though :/

sorry cant be, look at Kfunk's... should be close to that. (or even that)
 

serge

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the way i see it, is you dont need to rotate it

Area of the slice= 1/2.Pi.r^2

r=1/2[ (4ax)^1/2 - 2/3x]... but that just gets ugly :(

then integrate the area for volume
 

KFunk

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haboozin said:
my answer is (around) ... half of yours

the answer they gave... much smaller.

i dont think they're right


This is what they do:

(R - r)^2

and it really bothers me, it should be R^2 - r^2
They're the curves y = 2&radic;(ax) and y = 2x/3 right? So the radius of each semi-circle is r = (&radic;(ax) - x/3).

The volume should then be &pi;/2 &sum; r<sup>2</sup> &delta;x from 0 to 9a as &delta;x-->0

= 2547a<sup>3</sup>π/20

Anything wrong with that reasoning?
 

KFunk

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serge said:
the way i see it, is you dont need to rotate it
Whether or not that method is a rotation is more of a conceptual issue.
 

serge

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post the book's answer (even if you think it's wrong)
 

serge

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KFunk said:
r = (&radic;(ax) - x/3).


Anything wrong with that reasoning?
i think, when you square your radius you're integral has more than
1 power of a?

might be doing it wrong
 

KFunk

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serge said:
i think, when you square your radius you're integral has more than
1 power of a?

might be doing it wrong
It works out I think:

ax + x<sup>2</sup>/9 - (2&radic;a/3)x<sup>3/2</sup> ... integrate

[(ax<sup>2</sup>)/2 + x<sup>3</sup>/27 - (4&radic;a/15)x<sup>5/2</sup>] from 0 to 9a.

[81a<sup>3</sup>/2 + 27a<sup>3</sup> - 972a<sup>3</sup>/15]

The limits include an 'a' term so everything evens out to leave a<sup>3</sup>.
 

KFunk

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And it looks like I can't add numbers together :D. My previously posted answer of 2547a<sup>2</sup>π/20 should probably be 27a<sup>3</sup>&pi;/20 , that is, if I even added it up correctly this time.
 

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