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

1. Coolsector

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2. Coolsector

just this part ?

3. JoãoVitorMC

i think i need to consider the negative x values too

4. Coolsector

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5. JoãoVitorMC

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6. JoãoVitorMC

this part rotated around the y-axis

7. Coolsector

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

so i think we should separate into positive x and negative x

9. Coolsector

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

agree so far?

11. JoãoVitorMC

yes

12. Coolsector

wait .. i think that we dont need the positive x part at all

13. Coolsector

since the revolution of the negative part will cover it

14. JoãoVitorMC

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

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

them after i calculate this two parts i sum them?

17. Coolsector

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

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

yeah, after the rotation it will cover the positive part

20. JoãoVitorMC

ok a got it! thank you!

21. Coolsector

so if we add the positive part we get an extra for the answer and its bad..

22. JoãoVitorMC

yeah... that question is fun rsrsr

23. Coolsector

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

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

(a-b)^2 = (b-a)^2

26. JoãoVitorMC

because the function 3-x² is in far of y = x in relation to the y-axis?

27. Coolsector

(a-b)^2 = (b-a)^2 !!

28. Coolsector

and anyway it's better (for me at least) to think of subtracting the small number from the high number

29. JoãoVitorMC

ohh ok i got it!

30. JoãoVitorMC

thank you!!

31. Coolsector

;)

32. Coolsector

oh wait!!

33. Coolsector

there is something wrong there!

34. Coolsector

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

sorry for that.. so long havent done such things

36. Coolsector

because we like subtracting the inner disc from the larger disc