## PhoenixFire Group Title 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. one year ago one year ago

1. e.mccormick Group Title

The limits would be the z axis change.

2. e.mccormick Group Title

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

3. e.mccormick Group Title

That would be one way to do it.

4. genius12 Group Title

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

5. PhoenixFire Group Title

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

6. e.mccormick Group Title

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 Group Title

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

8. e.mccormick Group Title

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

9. PhoenixFire Group Title

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

10. e.mccormick Group Title

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

11. zugzwang Group Title

lol hoa dont bring real analysis into integrals confusing!

12. PhoenixFire Group Title

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

13. zugzwang Group Title

no, i'm still studying this

14. zugzwang Group Title

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

15. zugzwang Group Title

polar coords okay?

16. zugzwang Group Title

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

17. PhoenixFire Group Title

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

18. zugzwang Group Title

guys u still there?

19. PhoenixFire Group Title

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

20. zugzwang Group Title

i dont have one. :(

21. zugzwang Group Title

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

22. PhoenixFire Group Title

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 Group Title

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. zugzwang Group Title

i have an idea

25. zugzwang Group Title

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

26. zugzwang Group Title

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

27. zugzwang Group Title

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

28. zugzwang Group Title

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

29. zugzwang Group Title

30. zugzwang Group Title

no i dont speak tieng viet

31. zugzwang Group Title

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

32. zugzwang Group Title

big brother/sister lol

33. zugzwang Group Title

dont know what that means

34. zugzwang Group Title

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

35. zugzwang Group Title

No. I'm a college freshman.

36. PhoenixFire Group Title

@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. zugzwang Group Title

then I gots no clue sorry :)

38. zugzwang Group Title

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

39. PhoenixFire Group Title

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

40. zugzwang Group Title

that works too

41. zugzwang Group Title

but yeah, it's not gonna be straight

42. zugzwang Group Title

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

43. PhoenixFire Group Title

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