## amistre64 5 years ago Find the volume of the solid created by the bounded region of (x=y-y^2) and (x=y^2 -3) when it is spun around the line (x=-4).

1. anonymous

Ah this question still o.O

2. amistre64

Yeah :) but I found out what I was doing wrong.... other than trying to drink and derive :)

3. anonymous

So we at least know you can add 4 to all terms to change the bounds and have it revolved around x = 0

4. anonymous

And it's a torus, right? So you need to integrate the area of the torus around the line in a full circle.

5. anonymous

A cross section of the torus.

6. amistre64

yep...

7. anonymous

Can't you just integrate it?

8. amistre64

its pi (S) [4+F(y)]^2 - [4+G(y)]^2 | -1, 3/2 I kept trying: pi (S) [(4+F(y)) - (4+G(y))]^2 | -1, 3/2

9. amistre64

math: yes, but couldnt figure out how to last night :)

10. anonymous

$x = y - y^2 + 4$ $x = y^2 + 1$

11. anonymous

Okay so the bounds are y = -3, y = 1/2

12. anonymous

No...

13. anonymous

What?

14. amistre64

Krb: goog good. the bounds of integration are -1 and 3/2

15. anonymous

Yeah those are the bounds ^

16. anonymous

i need to figure out how to use the equations to find the solids of revolution..if this is x=-4, then the line we get is a vertical one right? so its been revolved upwards?

17. amistre64

you got the equation right; now square them individually before you subtract them..

18. anonymous

So $\int\limits_{(-1)}^{(3/2)}(y - y^2 + 4) - (y^2 + 1)$

19. anonymous

That's what i would think

20. anonymous

Krb you're always on this site. I like how when I look at paul's notes, you were the first one on open study.

21. amistre64

think; the rotation is around the "y axis" so to speak

22. anonymous

yeah..thats what i mean, so its like integrating the difference of squares of the equations right? and these equations signify -----?

23. amistre64

Krb: getting closer. we need to integrate [F(y)]^2 - [G(x)]^2 in the bounds

24. anonymous

Yes, normally when integrating with respect to x you are finding the are underneath the curve that drops to the x axis. So in this case you are finding the area "to the side of the curve" that drops to the y axis

25. amistre64

Think; exactly :)

26. anonymous

I'm confused as to why it would be squared

27. amistre64

typo...[G(y)]^2

28. anonymous

i guess it should be squared...shouldn't it?

29. amistre64

Krb: if we do it seperately for each on, there is no real "area" so to speak until we combine them. Does that make sense?

30. amistre64

pi* (S) F(y)^2 dy - (S) G(y)^2 dy The equations ARE our Radiusessess

31. anonymous

OH

32. anonymous

if y1 & y2 are the curves then , $\int\limits_{}^{}[y1^{2}-y2^{2}]$ gives the volume of the solid? where y1=upper curve & y2 the lower

33. anonymous

multiplied by a pi

34. anonymous

SO we are looking at it from a top down view, and then you need pi * r^2 for both "circles"

35. amistre64

exactly

36. anonymous

on efor inner and the other for outer 'circle' i guess

37. amistre64

think: thats the gyst of it yeah

38. anonymous

thnx!!

39. anonymous

$[\pi \int\limits_{-1}^{(3/2)}(y - y^2 + 4)^2] - [\pi \int\limits_{-1}^{(3/2)}(y^2 + 1)^2]$

40. amistre64

41. anonymous

No, you are :)

42. anonymous

u both are :)

43. amistre64

we should start a mensa club right here lol

44. anonymous

the open study math mensa

45. anonymous

:) :) so krb ur'e pursuing your masters too?

46. amistre64

C1 - C2 is what we were missing last night. 2pi(R) - 2pi(r) does not equal 2pi(R-r)

47. amistre64

it does, but it doesnt when you integrate lol

48. anonymous

wer those constants that u'll missed out yesterday?

49. anonymous

Ends up being 85.9029

50. anonymous

think: I'm honestly not sure

51. anonymous

krb: that's alright

52. anonymous

i dunno why it gets posted twice always, when i click only once!

53. anonymous

I think I probably would like to get a Masters

54. anonymous

good