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

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

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

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

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

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

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.1ok 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)

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

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.1for 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

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.1wait .. i think that we dont need the positive x part at all

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

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

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.1so 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

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

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.1wait 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
 2 years ago
Best ResponseYou've already chosen the best response.1i 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

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

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

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.1so if we add the positive part we get an extra for the answer and its bad..

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

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.1lol 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 ?

JoãoVitorMC
 2 years ago
Best ResponseYou've already chosen the best response.0only 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 ?

JoãoVitorMC
 2 years ago
Best ResponseYou've already chosen the best response.0because the function 3x² is in far of y = x in relation to the yaxis?

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

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

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

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.1integral 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

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

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