# Boundary values for integral

1. Aug 8, 2011

### Inertigratus

1. The problem statement, all variables and given/known data
Find the volume between z = x2 + y2 and z = 2 - (x2 + y2).

2. Relevant equations

3. 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\int$1 dxdydz, after switching to polar coordinates I get
$\int\int\int$r drd$\Theta$dz

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?

2. Aug 8, 2011

### 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.

3. Aug 8, 2011

### tiny-tim

Hi Inertigratus!

(have a square-root: √ and a pi: π and a ≤ )
nooo , that's the whole of the volume

4. Aug 9, 2011

### 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

5. Aug 9, 2011

### HallsofIvy

Staff Emeritus
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 $z= r^2$.

6. Aug 9, 2011

### 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 in to volumes.
Above z = 1 and below z = 1.