# Function for triple integral

1. Mar 13, 2009

### -EquinoX-

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

http://img5.imageshack.us/img5/5222/53026504.th.jpg [Broken]

2. Relevant equations

3. The attempt at a solution

I know A-F except for what E is here, I answered sqrt(x^2+y^2) but it is wrong, so what is it supposed to be?

Last edited by a moderator: May 4, 2017
2. Mar 13, 2009

### bob1182006

You need to convert $\frac{1}{\sqrt{x^2+y^2+z^2}}$ into an equation that uses only rho, phi, and theta.

You have covered some formulas from going from Cartesian to Spherical coordinates, one of those applies to this function.

3. Mar 13, 2009

### -EquinoX-

sorry I posted the wrong question, please check now

4. Mar 13, 2009

### bob1182006

You need to convert x^2+y^2 to polar coordinates. I think that's your only problem.

5. Mar 13, 2009

### -EquinoX-

and how can I do that? is it just x^2+y^2?

all I know about polar coordinate is:
x = r cos theta
y = r sin theta
z = z

Last edited: Mar 13, 2009
6. Mar 13, 2009

### bob1182006

x^2+y^2 = ? You should have covered some formulas relating r to x^2+y^2 for polar coordinates.
You replace x^2+y^2 by that and then you find your lower limit for z.

7. Mar 13, 2009

### -EquinoX-

hmm..so the lower bound is r?

Last edited: Mar 13, 2009
8. Mar 13, 2009

### bob1182006

No, your original function is sqrt(x^2+y^2) so you'd have sqrt(r^2).

9. Mar 13, 2009

### -EquinoX-

sqrt r^2 is r right?

10. Mar 13, 2009

### bob1182006

Yes, well -r or r, but it's convention that r is always positive, so yes, r is the lower limit.