Volume Between Two Surfaces Defined by Boundary Values

[Inertigratus]

Homework Statement

Find the volume between z = x² + y² and z = 2 - (x² + y²).

Homework Equations

The Attempt at a Solution

if r² = x² + y²
then the lower part of the volume is defined by:
r² \leq z \leq 2 - r²
and: 0 \leq r \leq 1
the upper part by:
2 - r² \leq z \leq r²
and: 1 \leq r \leq \sqrt{2}

\int\int\int1 dxdydz, after switching to polar coordinates I get
\int\int\intr drd\Thetadz

Theta varies from 0 to 2 pi. That leaves me with taking the integral with respect to r and z.
I do it for z first, then finally for r. Then add the two volumes. But it's wrong.

Any ideas?

 

[cragar]

those 2 curves intersect at x=1 and the plane z=1 , I would break it up into 2 parts and go from z=1 to the top of the paraboloid and then do the second part from the bottom paraboloid to the plane z=1 , because if you don't do that your radius is not always going to the same thing. Actually now that I look at it you kinda did that.

 

[tiny-tim]

Hi Inertigratus!

(have a square-root: √ and a pi: π and a ≤ )

Find the volume between z = x² + y² and z = 2 - (x² + y²).
…
… the lower part of the volume is defined by:
r² \leq z \leq 2 - r²
and: 0 \leq r \leq 1

nooo , that's the whole of the volume

 

[Inertigratus]

Hehe, now you got me confused... :P
It's the whole volume?
I split it, because when r = 1, then 1 ≤ z ≤ 1, if you go up to r = √(2) then 2 ≤ z ≤ 0, which looks a bit weird since 0 is not greater than 2...
So lower part from 0 ≤ r ≤ 1, upper part from 1 ≤ r ≤ √(2)
Is that wrong? :P

 

[HallsofIvy]

I guess the question is: what do you mean by the "lower part"? What you have, 2- r^2\le z\le r^2, is incorrect because the paraboloid z= 2- r^2 is above the parabolid z= r^2.

 

[Inertigratus]

It's confusing...
Because, I know that when r = 0, then 2−r² is obviously larger than r².
What's confusing me is that when r is increasing, and reaches r² = 2, then the inequality doesn't make any sense. 2 ≤ z ≤ 0 makes no sense.
So I was thinking as I said in my first post, that I would have to split it into volumes.
Above z = 1 and below z = 1.