# Double integral, cylindrical coordinates

1. Nov 14, 2013

### Ed Aboud

1. The problem statement, all variables and given/known data

The problem states: Use cylindrical coordinates to evaluate

$\iiint_V \sqrt{x^2 +y^2 +z^2} \,dx\,dy\,dz$

where V is the region bounded by the plane $z = 3$ and the cone $z = \sqrt{x^2 + y^2}$

2. Relevant equations
$x = r cos( \theta )$
$y = r sin( \theta )$
$z = z$

$dV = dx dy dz = r dz dr d \theta$

3. The attempt at a solution

Changing to cylindrical coordinates:

$\iiint_V r \sqrt{r^2 +z^2} \,dz\,dr\,d \theta$

The limits are:

$3 \le z \le r$
$0 \le r \le 9$ ???
$0 \le \theta \le 2 \pi$

I'm not sure how to tackle this integral. Attempting to evaluate it in mathematica returns an error too. To me, this question would be easier to solve using spherical polar coordinates, but the question states cylindrical.

One thing to note, $\sqrt{r^2 +z^2} = R$ is the equation for a sphere of radius R, in cylindrical coordinates. Not sure if this may play a part in the solution.

Any help would be greatly appreciated.

2. Nov 14, 2013

### vanhees71

First think about your boundaries! In your cylinder coordinates they are given by the plane $z=3$ and the cone $z=r$. Thus for each $z$, from where to where runs $r$? What's the maximal value of $r$? Note that first integrating over $r$ and then over $z$ is simpler than the other order!

3. Nov 14, 2013

### Ed Aboud

So am I correct in saying:

$3 \le r \le 9$ ?

Correction:

$3 \le r \le z$ ?

4. Nov 14, 2013

### Ed Aboud

Apologies but I still can't get it to work out. My limits are:

$3 \le r \le z$
$3 \le z \le r$
$0 \le \theta \le 2 \pi$

I'm fairly sure this is wrong, but can't figure out the correct ones

5. Nov 14, 2013

### vanhees71

Just make a drawing to show you the region. Than it's easier to read it off. As far as I understand the question, the region should be the interior of a cone of height 3.

6. Nov 14, 2013

### Ed Aboud

So I plotted it out, and by inspection I concluded that the limits are:

$0 \le r \le 3$
$0 \le z \le 3$
$0 \le \theta \le 2 \pi$

However, I can't get past the integral:

$\iiint_V r \sqrt{r^2 + z^2} \,dr\,dz\,d \theta$

gives

$\iint_V \frac{1}{3} (r^2 + z^2)^(3/2) \,dz\,d \theta$

integrating this out i got really odd answers with inverse hyperbolic sines so I'm guessing it's probably wrong

7. Nov 14, 2013

### LCKurtz

Don't those limits describe a cylinder, not a cone?