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JoãoVitorMC

  • 3 years ago

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

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  1. Coolsector
    • 3 years ago
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    |dw:1349809459606:dw|

  2. Coolsector
    • 3 years ago
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    just this part ?

  3. JoãoVitorMC
    • 3 years ago
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    i think i need to consider the negative x values too

  4. Coolsector
    • 3 years ago
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    |dw:1349809625381:dw|

  5. JoãoVitorMC
    • 3 years ago
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    |dw:1349809619079:dw|

  6. JoãoVitorMC
    • 3 years ago
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    this part rotated around the y-axis

  7. Coolsector
    • 3 years ago
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    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)

  8. Coolsector
    • 3 years ago
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    so i think we should separate into positive x and negative x

  9. Coolsector
    • 3 years ago
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    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

  10. Coolsector
    • 3 years ago
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    agree so far?

  11. JoãoVitorMC
    • 3 years ago
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    yes

  12. Coolsector
    • 3 years ago
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    wait .. i think that we dont need the positive x part at all

  13. Coolsector
    • 3 years ago
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    since the revolution of the negative part will cover it

  14. JoãoVitorMC
    • 3 years ago
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    i was thinking in use \[\int\limits_{0.5(-1-\sqrt(13))}^{0.5(\sqrt(13)-1)}2\pi x (-x² -x +3) dx\]

  15. Coolsector
    • 3 years ago
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    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

  16. JoãoVitorMC
    • 3 years ago
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    them after i calculate this two parts i sum them?

  17. Coolsector
    • 3 years ago
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    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"

  18. Coolsector
    • 3 years ago
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    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

  19. JoãoVitorMC
    • 3 years ago
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    yeah, after the rotation it will cover the positive part

  20. JoãoVitorMC
    • 3 years ago
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    ok a got it! thank you!

  21. Coolsector
    • 3 years ago
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    so if we add the positive part we get an extra for the answer and its bad..

  22. JoãoVitorMC
    • 3 years ago
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    yeah... that question is fun rsrsr

  23. Coolsector
    • 3 years ago
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    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 ?

  24. JoãoVitorMC
    • 3 years ago
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    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 ?

  25. Coolsector
    • 3 years ago
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    (a-b)^2 = (b-a)^2

  26. JoãoVitorMC
    • 3 years ago
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    because the function 3-x² is in far of y = x in relation to the y-axis?

  27. Coolsector
    • 3 years ago
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    (a-b)^2 = (b-a)^2 !!

  28. Coolsector
    • 3 years ago
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    and anyway it's better (for me at least) to think of subtracting the small number from the high number

  29. JoãoVitorMC
    • 3 years ago
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    ohh ok i got it!

  30. JoãoVitorMC
    • 3 years ago
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    thank you!!

  31. Coolsector
    • 3 years ago
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    ;)

  32. Coolsector
    • 3 years ago
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    oh wait!!

  33. Coolsector
    • 3 years ago
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    there is something wrong there!

  34. Coolsector
    • 3 years ago
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    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

  35. Coolsector
    • 3 years ago
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    sorry for that.. so long havent done such things

  36. Coolsector
    • 3 years ago
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    because we like subtracting the inner disc from the larger disc

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