anonymous
  • anonymous
Find the volume generated by revolution of the region y = x and y = 3-x², around the y-axis
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1349809459606:dw|
anonymous
  • anonymous
just this part ?
anonymous
  • anonymous
i think i need to consider the negative x values too

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

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