JoãoVitorMC
Find the volume generated by revolution of the region y = x and y = 3-x², around the y-axis
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Coolsector
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|dw:1349809459606:dw|
Coolsector
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just this part ?
JoãoVitorMC
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i think i need to consider the negative x values too
Coolsector
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|dw:1349809625381:dw|
JoãoVitorMC
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|dw:1349809619079:dw|
JoãoVitorMC
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this part rotated around the y-axis
Coolsector
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ok so intersection of y=x and y=3-x^2 is
3 -x^2 =x
x^2 +x -3 = 0
x = 0.5(-1-sqrt(13))
x = 0.5(sqrt(13)-1)
Coolsector
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so i think we should separate into positive x and negative x
Coolsector
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for the positive x there are two parts :
integral of pi*(y)^2dy from y =0 to y = 0.5(sqrt(13)-1)
and
integral of pi*(sqrt(y-3))^2dy from y = 0.5(sqrt(13)-1) to y= 3
Coolsector
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agree so far?
JoãoVitorMC
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yes
Coolsector
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wait .. i think that we dont need the positive x part at all
Coolsector
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since the revolution of the negative part will cover it
JoãoVitorMC
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i was thinking in use \[\int\limits_{0.5(-1-\sqrt(13))}^{0.5(\sqrt(13)-1)}2\pi x (-x² -x +3) dx\]
Coolsector
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so i would go only for the negative part..
there are two parts :
integral of pi*(y-(-sqrt(y-3)))^2*dy from y= 0.5(-1-sqrt(13)) to y = 0
and
integral of pi*(sqrt(y-3))^2*dy from y = 0 to y = 3
JoãoVitorMC
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them after i calculate this two parts i sum them?
Coolsector
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wait but dont you agree with: " i think that we dont need the positive x part at all
since the revolution of the negative part will cover it"
Coolsector
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i would only add those 2 :
integral of pi*(y-(-sqrt(y-3)))^2*dy from y= 0.5(-1-sqrt(13)) to y = 0
and
integral of pi*(sqrt(y-3))^2*dy from y = 0 to y = 3
JoãoVitorMC
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yeah, after the rotation it will cover the positive part
JoãoVitorMC
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ok a got it! thank you!
Coolsector
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so if we add the positive part we get an extra for the answer and its bad..
JoãoVitorMC
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yeah... that question is fun rsrsr
Coolsector
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lol
so you agree with :
integral of pi*(y-(-sqrt(y-3)))^2*dy from y= 0.5(-1-sqrt(13)) to y = 0
and
integral of pi*(sqrt(y-3))^2*dy from y = 0 to y = 3
?
JoãoVitorMC
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only one thing is strange, sholdn't it be
integral of pi*((-sqrt(y-3))-y)^2*dy from y= 0.5(-1-sqrt(13)) to y = 0
and
integral of pi*(sqrt(y-3))^2*dy from y = 0 to y = 3
?
Coolsector
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(a-b)^2 = (b-a)^2
JoãoVitorMC
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because the function 3-x² is in far of y = x in relation to the y-axis?
Coolsector
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(a-b)^2 = (b-a)^2 !!
Coolsector
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and anyway it's better (for me at least) to think of subtracting the small number from the high number
JoãoVitorMC
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ohh ok i got it!
JoãoVitorMC
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thank you!!
Coolsector
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;)
Coolsector
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oh wait!!
Coolsector
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there is something wrong there!
Coolsector
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integral of pi*(y-(-sqrt(y-3)))^2*dy from y= 0.5(-1-sqrt(13)) to y = 0
i should square them both -> y-(-sqrt(y-3))
we should do :
integral of pi*[(sqrt(y-3))^2 - y^2]*dy from y= 0.5(-1-sqrt(13)) to y = 0
Coolsector
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sorry for that.. so long havent done such things
Coolsector
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because we like subtracting the inner disc from the larger disc