## Dido525 one year ago Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = -5. y=x^2 , x=y^2

1. Dido525

Anyone want to help me set up the intergal? I can do the rest.

2. exraven

sketch the region first

3. Dido525

|dw:1358400866711:dw|I did:

4. Dido525

Something like that roughly.

5. exraven

can you determine the radius and the height?

6. Dido525

No I can't actually lol.

7. Dido525

Would the rdiube 5?

8. Dido525

9. exraven

|dw:1358400896304:dw|

10. exraven

the radius and the height must be a function of y

11. Dido525

Right.

12. Dido525

So y^2-(-5)

13. exraven

yes, and the height?

14. Dido525

1?

15. exraven

the height is the rightmost curve minus the leftmost curve right?

16. exraven

the height must also be a function of y

17. Dido525

|dw:1358401360267:dw|

18. Dido525

Ohh....I see....

19. exraven

actually its not like that, wait ill draw again

20. Dido525

Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.

21. exraven

|dw:1358401483221:dw|

22. Dido525

Yeah, that's better.

23. exraven

|dw:1358401575777:dw|

24. Dido525

Right - left the I suppose right?

25. Dido525

x^2-sqrt(x) .

26. exraven

yes, but it must be a function of y

27. Dido525

Why not x?

28. Dido525

Same function regardless right?

29. exraven

because we integrate it with respect to y later

30. Dido525

Yeah I see it now haha.

31. Dido525

so sqrt(y)-y^2?

32. exraven

yes

33. exraven

and here is the way to determine the radius|dw:1358402147166:dw| the radius is y + 5

34. Dido525

the limits would be from 0 to 1 I believe or am I wrong?

35. Dido525

The radius would simply be the parbola - the line.

36. Dido525

+ the line sorry.

37. exraven

$\Delta V = 2 \pi R h \Delta y$$\Delta V = 2 \pi (y + 5)(\sqrt{y} - y^2) \Delta y$$V = \int\limits_{0}^{1} 2 \pi (y + 5)(\sqrt{y} - y^2) dy$

38. Dido525

1 sec,I need to process that.

39. Dido525

Woudn't it b sqrt(y)+5?

40. Dido525

be*

41. exraven

this is for the radius|dw:1358403009035:dw|this is for the height|dw:1358403125789:dw|

42. Dido525

The answer I got was 2 which is wrong.

43. Dido525

Nvm. It's right.

44. Dido525

109pi/30

45. exraven

feel free to ask if its hard to follow

46. Dido525

Why isn't the height sqrt(x)+5?

47. exraven

it must be a function of y, because we integrate it with respect to y, sqrt(x) + 5 = y + 5

48. exraven

and it is not the height, it is the radius

49. exraven

the radius is sqrt(y) - y^2

50. Dido525

y^2 + 5 sorry.

51. Dido525

Why not that?

52. exraven

|dw:1358403804913:dw|

53. Dido525

Right. That's root x.

54. Dido525

Ahh, that distance is y and add 5.

55. exraven

yup, its sqrt(x), because the curve is x = y^2 sqrt(x) = sqrt(y^2) = y

56. Dido525

What's dy?

57. Dido525

Thickness right?

58. exraven

yes, because the cylinder looks like this|dw:1358404151779:dw|we integrate it with respect to y, because we fill the cylinder downward

59. Dido525

Thanks so much!

60. exraven

here is more details, I am sorry if I cannot explain further, because I teach myself all this stuff http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

61. Dido525

Thanks :) .

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