## JoãoVitorMC Group Title Find the volume generated by revolution of the region y = x and y = 3-x², around the y-axis one year ago one year ago

1. Coolsector Group Title

|dw:1349809459606:dw|

2. Coolsector Group Title

just this part ?

3. JoãoVitorMC Group Title

i think i need to consider the negative x values too

4. Coolsector Group Title

|dw:1349809625381:dw|

5. JoãoVitorMC Group Title

|dw:1349809619079:dw|

6. JoãoVitorMC Group Title

this part rotated around the y-axis

7. Coolsector Group Title

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 Group Title

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

9. Coolsector Group Title

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 Group Title

agree so far?

11. JoãoVitorMC Group Title

yes

12. Coolsector Group Title

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

13. Coolsector Group Title

since the revolution of the negative part will cover it

14. JoãoVitorMC Group Title

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 Group Title

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 Group Title

them after i calculate this two parts i sum them?

17. Coolsector Group Title

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 Group Title

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 Group Title

yeah, after the rotation it will cover the positive part

20. JoãoVitorMC Group Title

ok a got it! thank you!

21. Coolsector Group Title

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

22. JoãoVitorMC Group Title

yeah... that question is fun rsrsr

23. Coolsector Group Title

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 Group Title

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 Group Title

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

26. JoãoVitorMC Group Title

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

27. Coolsector Group Title

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

28. Coolsector Group Title

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

29. JoãoVitorMC Group Title

ohh ok i got it!

30. JoãoVitorMC Group Title

thank you!!

31. Coolsector Group Title

;)

32. Coolsector Group Title

oh wait!!

33. Coolsector Group Title

there is something wrong there!

34. Coolsector Group Title

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 Group Title

sorry for that.. so long havent done such things

36. Coolsector Group Title

because we like subtracting the inner disc from the larger disc