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PhoenixFire
 3 years ago
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
 3 years ago
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.

This Question is Closed

e.mccormick
 3 years ago
Best ResponseYou've already chosen the best response.0The limits would be the z axis change.

e.mccormick
 3 years ago
Best ResponseYou've already chosen the best response.0Then you rotate around y, with a floating radius of x.

e.mccormick
 3 years ago
Best ResponseYou've already chosen the best response.0That would be one way to do it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is that z = x^2 + y^2  1? @PhoenixFire

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, I don't know why they put the ( ) there. it's the same as \[z=x^2+y^21\]

e.mccormick
 3 years ago
Best ResponseYou've already chosen the best response.0It 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
 3 years ago
Best ResponseYou've already chosen the best response.0@e.mccormick I'm not quite sure how you would do that.

e.mccormick
 3 years ago
Best ResponseYou've already chosen the best response.0Or I think you can... hehe. I have been known to be wrong!

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0@Hoa Yes! Give us some direction if you know how to do it.

e.mccormick
 3 years ago
Best ResponseYou've already chosen the best response.0@Hoa Please do, I am checking my statement! And I may be completely off here.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol hoa dont bring real analysis into integrals confusing!

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0@Hoa I think polar coordinates might be the way to do this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, i'm still studying this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im just a chess player lol :D but okay let's see...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it looks less menacing that way \[\large z = r^2  1\]

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Maybe it is a triple integral. I thought it would be double.

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Yup, just looking through my textbook to see if maybe it is a Triple. But keep going with your solution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe it does have to be in triple integrals...

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0No 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
 3 years ago
Best ResponseYou've already chosen the best response.0Alright 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how about \[\Large \int\limits_{\theta = 0}^{2\pi}\int\limits_{r=1}^{\sqrt 2}(r^21) \ rdrd\theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if r goes from 1 to square root of 2, then z would go from 0 to 1.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im sorry i didn't see besides you probably deleted it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes we meet lol hi im Kevin Nguyen and you are...?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0american, vietnamese dad

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no i dont speak tieng viet

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you gave one, i gave it again only wait for his reaction are you a high school student?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0big brother/sister lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dont know what that means

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe I will you better not have cussed at me tho lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No. I'm a college freshman.

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then I gots no clue sorry :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, why does it have to be straight? it's not gonna be straight, it's an elliptic paraboloid for heavens sakes

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Can I not find the area of this and then rotate it around Z from 0 > 2pi? dw:1367636933452:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but yeah, it's not gonna be straight

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the straight (like a cone) graph would be \[\large z = r\]or \[\large z = \sqrt{x^2+y^2}\]

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, it's definitely not going to be straight.
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