PhoenixFire
  • PhoenixFire
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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e.mccormick
  • e.mccormick
The limits would be the z axis change.
e.mccormick
  • e.mccormick
Then you rotate around y, with a floating radius of x.
e.mccormick
  • e.mccormick
That would be one way to do it.

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More answers

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

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