the mother of all p2 volumes of rev. question

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hihihihi
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Find the volume when the following shaded region is rotated 360 degrees about the y-axis.


Book says answer= pi.


well i worked out the vol of shaded region on one side of the y axis- do i need to double this or is this the final answer?? because the shaded is symetrical about yaxis, and surely the same volume is achieved by not doubling??
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Lithium
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(Original post by hihihihi)
Find the volume when the following shaded region is rotated 360 degrees about the y-axis.


Book says answer= pi.


well i worked out the vol of shaded region on one side of the y axis- do i need to double this or is this the final answer?? because the shaded is symetrical about yaxis, and surely the same volume is achieved by not doubling??


Integral [ 2-x^-2-x^2] between the point of intersection....
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Magician
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omfg!! i would guess that it should equal half of volume... but how do you calculate it in the first place.???? Could you give a step-by-step please...!!
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hihihihi
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(Original post by Mathemagician)
omfg!! i would guess that it should equal half of volume... but how do you calculate it in the first place.???? Could you give a step-by-step please...!!
try looking at it side ways

get x in terms of y, so that you can integrate x with respect to y.
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lgs98jonee
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(Original post by hihihihi)
Find the volume when the following shaded region is rotated 360 degrees about the y-axis.


Book says answer= pi.


well i worked out the vol of shaded region on one side of the y axis- do i need to double this or is this the final answer?? because the shaded is symetrical about yaxis, and surely the same volume is achieved by not doubling??
volume=pi[INT(2-y) (between 1 and 2)+INTy (between 0 and 1)]
=pi((2y-y^2/2)+(y^2/2))
=pi(4-2-2+1/2+1/2)
=pi!!!
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friday13th
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im glad this didnt come up on AQA pure 2!!!
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hihihihi
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(Original post by lgs98jonee)
volume=pi[INT(2-y) (between 1 and 2)+INTy (between 0 and 1)]
=pi((2y-y^2/2)+(y^2/2))
=pi(4-2-2+1/2+1/2)
=pi!!!
so you dont need to double? :eek:
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lgs98jonee
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(Original post by hihihihi)
so you dont need to double? :eek:
well no cos it is symmetrical and so the other side goes through the same volume as the right hand side so no doubling
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Magician
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I got it!!!!!!!! vol = pi*int(2-x^2)^2 = 19/3pi - 10/3pi = 3pi (limits 1 and -1)

other volume = pi*int(x^2)^2 = pi (limits are 0 and -1). Do this twice as it is a reflection (limits 1 and 0). Add these together to get 2pi. (if done with limits 1 and -1, they will cancel each other out to give 0)

so volume shaded = 3pi-2pi
= pi
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lexazver203
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(Original post by Mathemagician)
I got it!!!!!!!! vol = pi*int(2-x^2)^2 = 19/3pi - 10/3pi = 3pi (limits 1 and -1)

other volume = pi*int(x^2)^2 = pi (limits are 0 and -1). Do this twice as it is a reflection (limits 1 and 0). Add these together to get 2pi. (if done with limits 1 and -1, they will cancel each other out to give 0)

so volume shaded = 3pi-2pi
= pi
wouldn you get x^5/5 when """other volume = pi*int(x^2)^2 = pi (limits are 0 and -1). Do this twice as it is a reflection (limits 1 and 0)""" therefore 1/5+1/5=2/5pi ??
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Lithium
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My book says its 16pi
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lexazver203
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(Original post by lgs98jonee)
volume=pi[INT(2-y) (between 1 and 2)+INTy (between 0 and 1)]
=pi((2y-y^2/2)+(y^2/2))
=pi(4-2-2+1/2+1/2)
=pi!!!
shouldnt you square f(x) when finding the volume of revolution?
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hihihihi
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(Original post by lexazver203)
shouldnt you square f(x) when finding the volume of revolution?
he did, but didnt write all the steps
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lexazver203
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i get it to be 28/5pi but i know this is aint right
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lexazver203
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(Original post by hihihihi)
he did, but didnt write all the steps
doesnt look so to me
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hihihihi
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split it into 2 sections

y= x²
=> x=√y
Vol. = Pi * ∫ (x²)dy
= pi * ∫ y dy
=pi*[ y²/2] between 1 and 0
etc.

y=2-x²
=> x=√(2-y)
Vol. = Pi * ∫ (x²)dy
= pi * ∫ (2-y) dy
= pi * [2y - y²/2] between 2 and 1.
etc.

i think that's right yeah?
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john !!
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y = x^2
y = 2 - x^2
x^2 = 2 - x^2
2x^2 = 2
x = +/- 1

Area = (1/-1){ (2 - x^2) dx - (1/1){ x^2 dx
= [2x - x^3/3](1/-1) - [x^3/3](1/-1)
= {[2*1 - 1^3/3] - [2*-1 - -1^3/3]} - {[1^3/3] - [-1^3/3]}
= {[10/6] - [2/3]}
= 1 square unit.

Square and multiply by pi:

pi.1^2 = pi cube units.
Great question!
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kimoni
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(Original post by mik1a)
y = x^2
y = 2 - x^2
x^2 = 2 - x^2
2x^2 = 2
x = +/- 1

Area = (1/-1){ (2 - x^2) dx - (1/1){ x^2 dx
= [2x - x^3/3](1/-1) - [x^3/3](1/-1)
= {[2*1 - 1^3/3] - [2*-1 - -1^3/3]} - {[1^3/3] - [-1^3/3]}
= {[10/6] - [2/3]}
= 1 square unit.

Square and multiply by pi:

pi.1^2 = pi cube units.
Great question!
This is driving me crazy... I'm trying to work through yours to see where I've gone wrong, but this little bit is driving me insane...

= {[2*1 - 1^3/3] - [2*-1 - -1^3/3]} - {[1^3/3] - [-1^3/3]}
= {[10/6] - [2/3]}

How the hell do you get 10/6 ? I've tried it over and over again, I get 10/3!!!

(2 - 1/3) - (-2 + 1/3) = 10/3
aaarrgh! whyyyyyyy?
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john !!
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(Original post by kimoni)
This is driving me crazy... I'm trying to work through yours to see where I've gone wrong, but this little bit is driving me insane...

= {[2*1 - 1^3/3] - [2*-1 - -1^3/3]} - {[1^3/3] - [-1^3/3]}
= {[10/6] - [2/3]}

How the hell do you get 10/6 ? I've tried it over and over again, I get 10/3!!!

(2 - 1/3) - (-2 + 1/3) = 10/3
aaarrgh! whyyyyyyy?
:confused:
I can't see myself... I guess it's a mistake.
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kimoni
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(Original post by mik1a)
:confused:
I can't see myself... I guess it's a mistake.
I don't think your method works.. I've even drawn up the graphs in a graphing program, which calculates the area under the curve. It tells me the area under x^2 is 2/3, and the area under the other is 10/3... so the area under the curve is 8/3

I don't think you can just square and times by pi... dammit. back to square 1

I don't see why i can't just do
(1/-1) pi INT: (2-x^2) - (x^2) dx
it doesnt work for me... argh
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