# Function for triple integral

#### -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?

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#### 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.

#### -EquinoX-

sorry I posted the wrong question, please check now

#### bob1182006

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

#### -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

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#### 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.

#### -EquinoX-

hmm..so the lower bound is r?

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#### bob1182006

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

#### -EquinoX-

sqrt r^2 is r right?

#### bob1182006

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

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