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.
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e.mccormick
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The limits would be the z axis change.
e.mccormick
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Then you rotate around y, with a floating radius of x.
e.mccormick
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That would be one way to do it.
genius12
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Is that z = x^2 + y^2 - 1?
@PhoenixFire
PhoenixFire
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Yeah, I don't know why they put the ( ) there.
it's the same as
\[z=x^2+y^2-1\]
e.mccormick
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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
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@e.mccormick I'm not quite sure how you would do that.
e.mccormick
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Or I think you can... hehe. I have been known to be wrong!
PhoenixFire
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@Hoa Yes! Give us some direction if you know how to do it.
e.mccormick
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@Hoa Please do, I am checking my statement! And I may be completely off here.
zugzwang
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lol hoa dont bring real analysis into integrals confusing!
PhoenixFire
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@Hoa I think polar coordinates might be the way to do this.
zugzwang
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no, i'm still studying this
zugzwang
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im just a chess player lol :D
but okay let's see...
zugzwang
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polar coords okay?
zugzwang
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it looks less menacing that way
\[\large z = r^2 - 1\]
PhoenixFire
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Maybe it is a triple integral. I thought it would be double.
zugzwang
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guys u still there?
PhoenixFire
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Yup, just looking through my textbook to see if maybe it is a Triple.
But keep going with your solution
zugzwang
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i dont have one. :(
zugzwang
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maybe it does have to be in triple integrals...
PhoenixFire
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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
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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
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i have an idea
zugzwang
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how about
\[\Large \int\limits_{\theta = 0}^{2\pi}\int\limits_{r=1}^{\sqrt 2}(r^2-1) \ rdrd\theta\]
zugzwang
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if r goes from 1 to square root of 2, then z would go from 0 to 1.
zugzwang
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im sorry i didn't see
besides you probably deleted it
zugzwang
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yes we meet lol
hi im Kevin Nguyen and you are...?
zugzwang
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american, vietnamese dad
zugzwang
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no i dont speak tieng viet
zugzwang
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you gave one, i gave it again only wait for his reaction
are you a high school student?
zugzwang
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big brother/sister lol
zugzwang
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dont know what that means
zugzwang
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maybe I will
you better not have cussed at me tho lol
zugzwang
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No. I'm a college freshman.
PhoenixFire
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@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
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then I gots no clue sorry :)
zugzwang
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wait, why does it have to be straight?
it's not gonna be straight, it's an elliptic paraboloid for heavens sakes
PhoenixFire
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Can I not find the area of this and then rotate it around Z from 0 -> 2pi?
|dw:1367636933452:dw|
zugzwang
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that works too
zugzwang
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but yeah, it's not gonna be straight
zugzwang
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the straight (like a cone) graph would be
\[\large z = r\]or
\[\large z = \sqrt{x^2+y^2}\]
PhoenixFire
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Yeah, it's definitely not going to be straight.