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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.

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

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

@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?

i dont have one. :(

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

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.

then I gots no clue sorry :)

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.