## PhoenixFire 3 years ago Double Integrals Find the volume of the solid $$z=(x^2 + y^2)-1$$ between the two plates z=0 and z=1 I'm not sure how to start. I've graphed the solid, but that hasn't helped me figure out what limits to use.

1. e.mccormick

The limits would be the z axis change.

2. e.mccormick

Then you rotate around y, with a floating radius of x.

3. e.mccormick

That would be one way to do it.

4. anonymous

Is that z = x^2 + y^2 - 1? @PhoenixFire

5. PhoenixFire

Yeah, I don't know why they put the ( ) there. it's the same as $z=x^2+y^2-1$

6. e.mccormick

It would be a shift on the z axis. You can still reduce it to a single axis problem, which was what I was thinking of.

7. PhoenixFire

@e.mccormick I'm not quite sure how you would do that.

8. e.mccormick

Or I think you can... hehe. I have been known to be wrong!

9. PhoenixFire

@Hoa Yes! Give us some direction if you know how to do it.

10. e.mccormick

@Hoa Please do, I am checking my statement! And I may be completely off here.

11. anonymous

lol hoa dont bring real analysis into integrals confusing!

12. PhoenixFire

@Hoa I think polar coordinates might be the way to do this.

13. anonymous

no, i'm still studying this

14. anonymous

im just a chess player lol :D but okay let's see...

15. anonymous

polar coords okay?

16. anonymous

it looks less menacing that way $\large z = r^2 - 1$

17. PhoenixFire

Maybe it is a triple integral. I thought it would be double.

18. anonymous

guys u still there?

19. PhoenixFire

Yup, just looking through my textbook to see if maybe it is a Triple. But keep going with your solution

20. anonymous

i dont have one. :(

21. anonymous

maybe it does have to be in triple integrals...

22. PhoenixFire

No I actually do not have answers. It's part of my assignment. We haven't learned triple integrals yet, so maybe I'm getting ahead of myself.

23. PhoenixFire

Alright guys, thanks for trying. I'm going to close this question and do some research into it. I'll come back and ask again in a few days if it's not sorted.

24. anonymous

i have an idea

25. anonymous

how about $\Large \int\limits_{\theta = 0}^{2\pi}\int\limits_{r=1}^{\sqrt 2}(r^2-1) \ rdrd\theta$

26. anonymous

if r goes from 1 to square root of 2, then z would go from 0 to 1.

27. anonymous

im sorry i didn't see besides you probably deleted it

28. anonymous

yes we meet lol hi im Kevin Nguyen and you are...?

29. anonymous

30. anonymous

no i dont speak tieng viet

31. anonymous

you gave one, i gave it again only wait for his reaction are you a high school student?

32. anonymous

big brother/sister lol

33. anonymous

dont know what that means

34. anonymous

maybe I will you better not have cussed at me tho lol

35. anonymous

No. I'm a college freshman.

36. PhoenixFire

@zugzwang That will give me the region |dw:1367636348171:dw| Right? Which isn't what we want. |dw:1367636411100:dw| This has to be a triple integral. I'll see if I understand it next week when we do triple integral.

37. anonymous

then I gots no clue sorry :)

38. anonymous

wait, why does it have to be straight? it's not gonna be straight, it's an elliptic paraboloid for heavens sakes

39. PhoenixFire

Can I not find the area of this and then rotate it around Z from 0 -> 2pi? |dw:1367636933452:dw|

40. anonymous

that works too

41. anonymous

but yeah, it's not gonna be straight

42. anonymous

the straight (like a cone) graph would be $\large z = r$or $\large z = \sqrt{x^2+y^2}$

43. PhoenixFire

Yeah, it's definitely not going to be straight.