## anonymous 5 years ago PLEASE SOLVE :) let R be the region bounded by the graph of y=x^2 -1 and the graph of x=y^2 .... find the volume of the solid generated when r is rotated about the vertical line x=2, and then write an integral expression that can be used to find the volume of the solid generated when R is rotated about the line y=-1

1. anonymous

I can get you started, but I may have to leave soon. If I help you set up the volume element, can you integrate from there?

2. anonymous

Well, I'll start anyway...

3. anonymous

I'm going to send through an image of the plots you need to look at.

4. anonymous

ok thanks!

5. anonymous

6. anonymous

this is hard without a calculator. I got the integral would be $\pi*\int\limits_{?}^{?}(2-y^2)^2-(2-\sqrt{y+1})^2dy$ and then i would need a calulator to get the limits where the two function are equal...is it supposed to have such nasty algebra or did I mess up soehwhere?

7. anonymous

Assuming your equations are correct, the shaded region is what you're looking at. I would plan on using cylindrical shells, and doing the integral in two parts: one where$y=\sqrt{x}$and one where$y=-\sqrt{x}$and then add the volumes up.

8. amistre64

plotting it was easy :) how we find the solutions? or is there another way?

9. anonymous

Actually, put cyl. shells on hold. When I first looked at your problem, I took the y-axis as a natural bound, but that's not the case since it's not in your question. When considering the region shaded, discs may be better. I want to work it out.

10. anonymous

I'm deriving it - it's easy to do on paper, but hard to explain. online

11. amistre64

deriving sounds interesting.... would you see where the slopes are perp to each other?

12. amistre64

been trying to solve it like system of equation....but no luck; then again, i could be missing something glaringly obvious :)

13. anonymous

the integral came out $\pi*[y^5/5-4y^3/3-y^2/2+8/3*y^(3/2)-y]...$

14. amistre64

jpick: you just subtract the equations from each other?

15. anonymous

but to get bounds for y i have to solve the values where y^4-y-1=0

16. amistre64

yeah.... been trying that one for hours :)

17. amistre64

just thinking outloud, but would newtons approximation work for us? or is that just to clumsy?

18. anonymous

Okay, it seems whatever method you use here (cylindrical shells or discs) the algebra is going to be crappy.

19. amistre64

loki... :) i agree

20. anonymous

Given that, I'm going to persist with cylindrical shells.

21. anonymous

maybe, some numerical approximation (or algebra solver program) will be needed

22. amistre64

finding the bounds is like pulling teeth, but without the lollipop at the end :)

23. amistre64

is it possible to set your own bounds and still get an answer?

24. anonymous

not sure what you mean? the bounds of integration are set by the intersection, or do you mean the approximation getting close enough will be our "own bounds"

25. amistre64

either way... I was wondering if going further than the bounds would have a canceling out effect, but most likely that is just wishful thinking

26. anonymous

Okay, the basic construction for the cylindrical shell is$\delta V = 2\pi(radius)(height)(elemental.thickness)$$=2\pi (x-2)(\sqrt{x}-(x^2-1)) \delta x$

27. anonymous

where...

28. anonymous

this element is for the section to the right of the line in the following image.

29. anonymous

30. anonymous

A similar situation will arise with the section to the left, where you have$\delta V=2\pi (2-x)(\sqrt{x}-(-\sqrt{x})) \delta x=2\pi(2-x)(2\sqrt{x}) \delta x$

31. anonymous

There's a mistake in the first element: x-2 should read as 2-x, the radius.

32. anonymous

You need to communicate with me. Is this working for you?

33. anonymous

Your limits of integration for the element I just derived above (the one with 2sqrt{x} for height) will be from x=0 to x=t where $-\sqrt{t}=t^2-1$ (i.e. get a number out for t).

34. anonymous

that will work, the only thing we need are the actual bounds, either in terms of x or y. I do have one question for you lokisan about the earlier differential equation thread. I got a different answer than you, even though we did it the same way given it was an exact equation

35. anonymous

The limits of the initial section I derived will be from x = the t you just found to x=s where $\sqrt{s}=s^2-1$(i.e. get an approximation for s).

36. anonymous

Which d.e.?

37. anonymous

The one where he wanted substitution?

38. anonymous

bethanymichalski, you there?

39. anonymous

yeah, I got x^2/2+y^/2+xy-x=c

40. anonymous

That's what I got using exact differential equation.

41. anonymous

Oh, i thought it said log(x+y)...my bad nm :(

42. anonymous

oh, he wanted to use a substitution method in the end. I didn't bother to check if the answers were the same, ;p