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

1. anonymous

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2. anonymous

just this part ?

3. anonymous

i think i need to consider the negative x values too

4. anonymous

|dw:1349809625381:dw|

5. anonymous

|dw:1349809619079:dw|

6. anonymous

this part rotated around the y-axis

7. 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)

8. anonymous

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

9. 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

10. anonymous

agree so far?

11. anonymous

yes

12. anonymous

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

13. anonymous

since the revolution of the negative part will cover it

14. anonymous

i was thinking in use $\int\limits_{0.5(-1-\sqrt(13))}^{0.5(\sqrt(13)-1)}2\pi x (-x² -x +3) dx$

15. 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

16. anonymous

them after i calculate this two parts i sum them?

17. 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"

18. 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

19. anonymous

yeah, after the rotation it will cover the positive part

20. anonymous

ok a got it! thank you!

21. anonymous

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

22. anonymous

yeah... that question is fun rsrsr

23. 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 ?

24. 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 ?

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

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

29. anonymous

ohh ok i got it!

30. anonymous

thank you!!

31. anonymous

;)

32. anonymous

oh wait!!

33. anonymous

there is something wrong there!

34. 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

35. anonymous

sorry for that.. so long havent done such things

36. anonymous

because we like subtracting the inner disc from the larger disc