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.

- PhoenixFire

- katieb

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- e.mccormick

The limits would be the z axis change.

- e.mccormick

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

- e.mccormick

That would be one way to do it.

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## More answers

- anonymous

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

- PhoenixFire

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

- 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.

- PhoenixFire

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

- e.mccormick

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

- PhoenixFire

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

- e.mccormick

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

- zugzwang

lol hoa dont bring real analysis into integrals confusing!

- PhoenixFire

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

- zugzwang

no, i'm still studying this

- zugzwang

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

- zugzwang

polar coords okay?

- zugzwang

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

- PhoenixFire

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

- zugzwang

guys u still there?

- PhoenixFire

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

- zugzwang

i dont have one. :(

- zugzwang

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

- 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.

- 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.

- zugzwang

i have an idea

- zugzwang

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

- zugzwang

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

- zugzwang

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

- zugzwang

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

- zugzwang

american, vietnamese dad

- zugzwang

no i dont speak tieng viet

- zugzwang

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

- zugzwang

big brother/sister lol

- zugzwang

dont know what that means

- zugzwang

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

- zugzwang

No. I'm a college freshman.

- 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.

- zugzwang

then I gots no clue sorry :)

- zugzwang

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

- PhoenixFire

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

- zugzwang

that works too

- zugzwang

but yeah, it's not gonna be straight

- zugzwang

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

- PhoenixFire

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

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