# the mother of all p2 volumes of rev. question

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Thread starter 15 years ago
#1
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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15 years ago
#2
(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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15 years ago
#3
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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Thread starter 15 years ago
#4
(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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15 years ago
#5
(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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15 years ago
#6
im glad this didnt come up on AQA pure 2!!!
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Thread starter 15 years ago
#7
(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?
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15 years ago
#8
(Original post by hihihihi)
so you dont need to double?
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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15 years ago
#9
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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15 years ago
#10
(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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15 years ago
#11
My book says its 16pi
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15 years ago
#12
(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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Thread starter 15 years ago
#13
(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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15 years ago
#14
i get it to be 28/5pi but i know this is aint right
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15 years ago
#15
(Original post by hihihihi)
he did, but didnt write all the steps
doesnt look so to me
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Thread starter 15 years ago
#16
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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15 years ago
#17
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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15 years ago
#18
(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?
0
15 years ago
#19
(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?

I can't see myself... I guess it's a mistake.
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15 years ago
#20
(Original post by mik1a)

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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