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

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

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

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

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

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

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
this part rotated around the yaxis
 2 years 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)
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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