Understanding an issue in a triple integral solution

Question

Here I have a triple integral

$$ \iiint f(x,y,z)dxdydz $$ on the region : $\{\sqrt[]{x^2+y^2} \le z \le \sqrt[]{4-x^2-y^2}\} $

if we use cylindrical coordinates we have : (1) $ r\le z \le \sqrt[]{4-r^2} $

and when we want to do the integral :

we determine that : $ 0 \le \theta \le 2\pi $ $ $ , $ $ $ 0 \le r \le \sqrt{2}$

my problem is the z

according the graph z is bigger than the Cone so it should be : between $0 \le z \le \sqrt{4-r^2}$
but according to equation I determine that z is between : $r \le z \le \sqrt{4-r^2}$

what I'm missing here ?

The cone opens up and the (hemi)sphere opens down, so the range for $z$ is above the cone ($z=r$) and below the sphere ($z=\sqrt{4-x^2-y^2}$). — rogerl, Jul 04 '14 at 20:12
But $r\le\sqrt{4-r^2}$ only if $r<\sqrt{2}$, so OP is correct. — Empy2, Jul 04 '14 at 20:13
@Nameless according to the concern and the graph I can see that we should write z starting from $0$ but why can't we deduce this from the equations — Waseem Francis, Jul 04 '14 at 20:15
If your limit of integration for $z$ goes from $0$ to $\sqrt{4 - r^2}$, you are not integrating over the cone then. — IAmNoOne, Jul 04 '14 at 20:16
so z can start from r or from $0$ but the integration way changes ? (but leads to the same answer) — Waseem Francis, Jul 04 '14 at 20:19
Maybe I am not fully understanding you, you say according the graph z is bigger than the Cone so it should be : between $0 \le z \le \sqrt{4-r^2}$ why? Why $z = 0$? — IAmNoOne, Jul 04 '14 at 20:23
The shape is bounded above by the sphere and below by the cone. — Mhenni Benghorbal, Jul 04 '14 at 20:25
@Nameless: I already paid attention to the triple integral. Thanks. — Mhenni Benghorbal, Jul 04 '14 at 20:28
I'm asking without relationg to the function. Why z starts from r and not from zero (according to the done equations) — Waseem Francis, Jul 04 '14 at 20:30
I did not I just asked if we get the same answer trying to find an answer to the discussed problem — Waseem Francis, Jul 04 '14 at 20:33
Ok , so from where should we start z from 0 ?(because of the graph) or from r (because of the written equations) and why ? Sorry I'm asking a lot — Waseem Francis, Jul 04 '14 at 20:37
If you start from $z = 0$, you are integrating over the hemisphere ($z = 0$ is a plane). — IAmNoOne, Jul 04 '14 at 21:59

mike · Accepted Answer · 2014-07-04T22:31:40.893

2

Here is the cross section of the integration volume in $r-z$ plane:

Here is the cross section in $r-z$ plane

The integration of the volume $D$ is given by (assuming that $f(x,y,z)=F(z,r)$): $$\int_D f(x,y,z)dxdydz=\int_0^{2\pi}d\theta \int_0^{\sqrt{2}} rdr \int_{r}^{\sqrt{4-r^2}}dz F(z,r)$$

edited Jul 04 '14 at 22:31

answered Jul 04 '14 at 20:51

mike

5,604

Thank you ! finally my problem is now almost cleared so to finish up can we do the integral in this order also ? :
$$\int_D f(x,y,z)dxdydz=\int_0^{\sqrt{2}}dr \int_0^{\sqrt{4-r^2}} dz \int_{0}^{2\pi}d\theta F(z,r)$$
– Waseem Francis Jul 04 '14 at 21:18
sure. $\int_0^{2\pi}d\theta=2\pi$ – mike Jul 04 '14 at 21:40

Understanding an issue in a triple integral solution

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