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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349809459606:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349809625381:dw

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i think we should separate into positive x and negative x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait .. i think that we dont need the positive x part at all

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since the revolution of the negative part will cover it

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait 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"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if we add the positive part we get an extra for the answer and its bad..

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol 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 ?

anonymous
 3 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 ?

anonymous
 3 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and anyway it's better (for me at least) to think of subtracting the small number from the high number

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there is something wrong there!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0integral 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry for that.. so long havent done such things

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because we like subtracting the inner disc from the larger disc
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