## anonymous 3 years 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. anonymous

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

2. anonymous

sketch the region first

3. anonymous

|dw:1358400866711:dw|I did:

4. anonymous

Something like that roughly.

5. anonymous

can you determine the radius and the height?

6. anonymous

No I can't actually lol.

7. anonymous

Would the rdiube 5?

8. anonymous

9. anonymous

|dw:1358400896304:dw|

10. anonymous

the radius and the height must be a function of y

11. anonymous

Right.

12. anonymous

So y^2-(-5)

13. anonymous

yes, and the height?

14. anonymous

1?

15. anonymous

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

16. anonymous

the height must also be a function of y

17. anonymous

|dw:1358401360267:dw|

18. anonymous

Ohh....I see....

19. anonymous

actually its not like that, wait ill draw again

20. anonymous

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

21. anonymous

|dw:1358401483221:dw|

22. anonymous

Yeah, that's better.

23. anonymous

|dw:1358401575777:dw|

24. anonymous

Right - left the I suppose right?

25. anonymous

x^2-sqrt(x) .

26. anonymous

yes, but it must be a function of y

27. anonymous

Why not x?

28. anonymous

Same function regardless right?

29. anonymous

because we integrate it with respect to y later

30. anonymous

Yeah I see it now haha.

31. anonymous

so sqrt(y)-y^2?

32. anonymous

yes

33. anonymous

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

34. anonymous

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

35. anonymous

The radius would simply be the parbola - the line.

36. anonymous

+ the line sorry.

37. anonymous

$\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. anonymous

1 sec,I need to process that.

39. anonymous

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

40. anonymous

be*

41. anonymous

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

42. anonymous

The answer I got was 2 which is wrong.

43. anonymous

Nvm. It's right.

44. anonymous

109pi/30

45. anonymous

feel free to ask if its hard to follow

46. anonymous

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

47. anonymous

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

48. anonymous

and it is not the height, it is the radius

49. anonymous

the radius is sqrt(y) - y^2

50. anonymous

y^2 + 5 sorry.

51. anonymous

Why not that?

52. anonymous

|dw:1358403804913:dw|

53. anonymous

Right. That's root x.

54. anonymous

Ahh, that distance is y and add 5.

55. anonymous

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

56. anonymous

What's dy?

57. anonymous

Thickness right?

58. anonymous

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

59. anonymous

Thanks so much!

60. anonymous

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. anonymous

Thanks :) .