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the mother of all p2 volumes of rev. question
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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??
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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#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??
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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#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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(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...!!
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.
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#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??
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!!!
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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!!!
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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#8
(Original post by hihihihi)
so you dont need to double?
so you dont need to double?
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#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
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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#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
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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#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!!!
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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(Original post by lexazver203)
shouldnt you square f(x) when finding the volume of revolution?
shouldnt you square f(x) when finding the volume of revolution?
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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?
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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#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!
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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#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!
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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#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?
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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#20
(Original post by mik1a)
I can't see myself... I guess it's a mistake.
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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