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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hihihihi
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- 20-06-2004 19:00
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- 20-06-2004 19:03
(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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- 20-06-2004 19:06
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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- 20-06-2004 19:08
(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...!!
get x in terms of y, so that you can integrate x with respect to y. -
lgs98jonee
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- 20-06-2004 19:15
(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??
=pi((2y-y^2/2)+(y^2/2))
=pi(4-2-2+1/2+1/2)
=pi!!! -
friday13th
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- 20-06-2004 19:17
im glad this didnt come up on AQA pure 2!!!
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hihihihi
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- 20-06-2004 19:18
(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!!! -
lgs98jonee
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- 20-06-2004 19:19
(Original post by hihihihi)
so you dont need to double? -
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- 20-06-2004 19:21
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 -
lexazver203
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- 20-06-2004 19:30
(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 -
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- 20-06-2004 19:32
My book says its 16pi
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lexazver203
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- 20-06-2004 19:35
(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!!! -
hihihihi
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- 20-06-2004 19:36
(Original post by lexazver203)
shouldnt you square f(x) when finding the volume of revolution? -
lexazver203
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- 20-06-2004 19:37
i get it to be 28/5pi but i know this is aint right
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lexazver203
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- 20-06-2004 19:41
(Original post by hihihihi)
he did, but didnt write all the steps -
hihihihi
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- 20-06-2004 19:47
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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- 20-06-2004 19:49
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! -
kimoni
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- 20-06-2004 20:23
(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!
= {[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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- 20-06-2004 20:27
(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. -
kimoni
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- 20-06-2004 20:33
(Original post by mik1a)
I can't see myself... I guess it's a mistake.
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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