Find the volume generated by revolution of the region y = x and y = 3-x², around the y-axis

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Find the volume generated by revolution of the region y = x and y = 3-x², around the y-axis

Mathematics
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|dw:1349809459606:dw|
just this part ?
i think i need to consider the negative x values too

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|dw:1349809625381:dw|
|dw:1349809619079:dw|
this part rotated around the y-axis
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)
so i think we should separate into positive x and negative x
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
agree so far?
yes
wait .. i think that we dont need the positive x part at all
since the revolution of the negative part will cover it
i was thinking in use \[\int\limits_{0.5(-1-\sqrt(13))}^{0.5(\sqrt(13)-1)}2\pi x (-x² -x +3) dx\]
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
them after i calculate this two parts i sum them?
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"
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
yeah, after the rotation it will cover the positive part
ok a got it! thank you!
so if we add the positive part we get an extra for the answer and its bad..
yeah... that question is fun rsrsr
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 ?
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 ?
(a-b)^2 = (b-a)^2
because the function 3-x² is in far of y = x in relation to the y-axis?
(a-b)^2 = (b-a)^2 !!
and anyway it's better (for me at least) to think of subtracting the small number from the high number
ohh ok i got it!
thank you!!
;)
oh wait!!
there is something wrong there!
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
sorry for that.. so long havent done such things
because we like subtracting the inner disc from the larger disc

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