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JoãoVitorMC
Group Title
Find the volume generated by revolution of the region y = x and y = 3x², around the yaxis
 one year ago
 one year ago
JoãoVitorMC Group Title
Find the volume generated by revolution of the region y = x and y = 3x², around the yaxis
 one year ago
 one year ago

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Coolsector Group TitleBest ResponseYou've already chosen the best response.1
dw:1349809459606:dw
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
just this part ?
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
i think i need to consider the negative x values too
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
dw:1349809625381:dw
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
dw:1349809619079:dw
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
this part rotated around the yaxis
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
ok so intersection of y=x and y=3x^2 is 3 x^2 =x x^2 +x 3 = 0 x = 0.5(1sqrt(13)) x = 0.5(sqrt(13)1)
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
so i think we should separate into positive x and negative x
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
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(y3))^2dy from y = 0.5(sqrt(13)1) to y= 3
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
agree so far?
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
wait .. i think that we dont need the positive x part at all
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
since the revolution of the negative part will cover it
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
i was thinking in use \[\int\limits_{0.5(1\sqrt(13))}^{0.5(\sqrt(13)1)}2\pi x (x² x +3) dx\]
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
so i would go only for the negative part.. there are two parts : integral of pi*(y(sqrt(y3)))^2*dy from y= 0.5(1sqrt(13)) to y = 0 and integral of pi*(sqrt(y3))^2*dy from y = 0 to y = 3
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
them after i calculate this two parts i sum them?
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
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"
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
i would only add those 2 : integral of pi*(y(sqrt(y3)))^2*dy from y= 0.5(1sqrt(13)) to y = 0 and integral of pi*(sqrt(y3))^2*dy from y = 0 to y = 3
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
yeah, after the rotation it will cover the positive part
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
ok a got it! thank you!
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
so if we add the positive part we get an extra for the answer and its bad..
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
yeah... that question is fun rsrsr
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
lol so you agree with : integral of pi*(y(sqrt(y3)))^2*dy from y= 0.5(1sqrt(13)) to y = 0 and integral of pi*(sqrt(y3))^2*dy from y = 0 to y = 3 ?
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
only one thing is strange, sholdn't it be integral of pi*((sqrt(y3))y)^2*dy from y= 0.5(1sqrt(13)) to y = 0 and integral of pi*(sqrt(y3))^2*dy from y = 0 to y = 3 ?
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
(ab)^2 = (ba)^2
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
because the function 3x² is in far of y = x in relation to the yaxis?
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
(ab)^2 = (ba)^2 !!
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
and anyway it's better (for me at least) to think of subtracting the small number from the high number
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
ohh ok i got it!
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
thank you!!
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
oh wait!!
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
there is something wrong there!
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
integral of pi*(y(sqrt(y3)))^2*dy from y= 0.5(1sqrt(13)) to y = 0 i should square them both > y(sqrt(y3)) we should do : integral of pi*[(sqrt(y3))^2  y^2]*dy from y= 0.5(1sqrt(13)) to y = 0
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
sorry for that.. so long havent done such things
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.1
because we like subtracting the inner disc from the larger disc
 one year ago
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