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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.

Mathematics
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The limits would be the z axis change.
Then you rotate around y, with a floating radius of x.
That would be one way to do it.

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Other answers:

Is that z = x^2 + y^2 - 1? @PhoenixFire
Yeah, I don't know why they put the ( ) there. it's the same as \[z=x^2+y^2-1\]
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.
@e.mccormick I'm not quite sure how you would do that.
Or I think you can... hehe. I have been known to be wrong!
@Hoa Yes! Give us some direction if you know how to do it.
@Hoa Please do, I am checking my statement! And I may be completely off here.
lol hoa dont bring real analysis into integrals confusing!
@Hoa I think polar coordinates might be the way to do this.
no, i'm still studying this
im just a chess player lol :D but okay let's see...
polar coords okay?
it looks less menacing that way \[\large z = r^2 - 1\]
Maybe it is a triple integral. I thought it would be double.
guys u still there?
Yup, just looking through my textbook to see if maybe it is a Triple. But keep going with your solution
i dont have one. :(
maybe it does have to be in triple integrals...
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.
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.
i have an idea
how about \[\Large \int\limits_{\theta = 0}^{2\pi}\int\limits_{r=1}^{\sqrt 2}(r^2-1) \ rdrd\theta\]
if r goes from 1 to square root of 2, then z would go from 0 to 1.
im sorry i didn't see besides you probably deleted it
yes we meet lol hi im Kevin Nguyen and you are...?
american, vietnamese dad
no i dont speak tieng viet
you gave one, i gave it again only wait for his reaction are you a high school student?
big brother/sister lol
dont know what that means
maybe I will you better not have cussed at me tho lol
No. I'm a college freshman.
@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.
then I gots no clue sorry :)
wait, why does it have to be straight? it's not gonna be straight, it's an elliptic paraboloid for heavens sakes
Can I not find the area of this and then rotate it around Z from 0 -> 2pi? |dw:1367636933452:dw|
that works too
but yeah, it's not gonna be straight
the straight (like a cone) graph would be \[\large z = r\]or \[\large z = \sqrt{x^2+y^2}\]
Yeah, it's definitely not going to be straight.

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