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PhoenixFire
Group Title
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.
 one year ago
 one year ago
PhoenixFire Group Title
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.
 one year ago
 one year ago

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e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
The limits would be the z axis change.
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
Then you rotate around y, with a floating radius of x.
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
That would be one way to do it.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Is that z = x^2 + y^2  1? @PhoenixFire
 one year ago

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

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
@e.mccormick I'm not quite sure how you would do that.
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
Or I think you can... hehe. I have been known to be wrong!
 one year ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
@Hoa Yes! Give us some direction if you know how to do it.
 one year ago

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

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
lol hoa dont bring real analysis into integrals confusing!
 one year ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
@Hoa I think polar coordinates might be the way to do this.
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
no, i'm still studying this
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
im just a chess player lol :D but okay let's see...
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
polar coords okay?
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
it looks less menacing that way \[\large z = r^2  1\]
 one year ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
Maybe it is a triple integral. I thought it would be double.
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
guys u still there?
 one year ago

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

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
i dont have one. :(
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
maybe it does have to be in triple integrals...
 one year ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
i have an idea
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
how about \[\Large \int\limits_{\theta = 0}^{2\pi}\int\limits_{r=1}^{\sqrt 2}(r^21) \ rdrd\theta\]
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
if r goes from 1 to square root of 2, then z would go from 0 to 1.
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
im sorry i didn't see besides you probably deleted it
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
yes we meet lol hi im Kevin Nguyen and you are...?
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
american, vietnamese dad
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
no i dont speak tieng viet
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
you gave one, i gave it again only wait for his reaction are you a high school student?
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
big brother/sister lol
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
dont know what that means
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
maybe I will you better not have cussed at me tho lol
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
No. I'm a college freshman.
 one year ago

PhoenixFire Group TitleBest 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.
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
then I gots no clue sorry :)
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
wait, why does it have to be straight? it's not gonna be straight, it's an elliptic paraboloid for heavens sakes
 one year ago

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

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
that works too
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
but yeah, it's not gonna be straight
 one year ago

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

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
Yeah, it's definitely not going to be straight.
 one year ago
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