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RivalryofTroll

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This question is for a friend:

How do I find the volume of this cross section? (HAVE TO USE SIMPSONS to find the area and then volume)

Refer to the below image for the question:
AxMG93ECIAEyuuk.jpg

Question in image, 'Find the volume of solid where x-axis is 12, and the cross sectional area is perpendicular'.
 
Last edited:

pheel

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I've never seen a question where they have given you the area, but haven't given you the formula.
Can't use similar area either , because it goes back down... or can you ? hmm, is it possible to derive the formula for the area ?
 

iSplicer

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This question is for a friend:

How do I find the volume of this cross section? (HAVE TO USE SIMPSONS to find the area and then volume)

Refer to the below image for the question:
View attachment 25840

Question in image, 'Find the volume of solid where x-axis is 12, and the cross sectional area is perpendicular'.
Haha, does your friend go to Truong?
 

pheel

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oh wow, iSplicer, how did you know?
its in the integration topic, and I never understood how to do it.
 

cineti970128

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hey i am not hundred sure if i am right but the whole point of Simpson's rule is to find area of the graph by section using parabola yes. then why not find the volume by section. then find the area of rotated cross section of graph with width ----> 0 which in this case
Well stuff this think about cross sectional piece which has radius A(x) as you mentioned thus cross sectional area = pie * A(x)^2 but you got 6 values of different A(x). there fore treat pie * A(x)^2 as G(x) and use simpson's rule on this substituted values 6 different G(x) to find the total volume.
Because integral of G(x) dx is the volume
i am pretty sure this should work

if not please correct me
 

iSplicer

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oh wow, iSplicer, how did you know?
its in the integration topic, and I never understood how to do it.
I went to him for almost five years, he was one my foremost mentors/role models. I could recognise simply by looking at the notes.

Tell Mr. Mai Satvik says hi!
 

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