Function for triple integral

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:
You need to convert [itex]\frac{1}{\sqrt{x^2+y^2+z^2}}[/itex] 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.
 
sorry I posted the wrong question, please check now
 
You need to convert x^2+y^2 to polar coordinates. I think that's your only problem.
 
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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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.
 
hmm..so the lower bound is r?
 
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No, your original function is sqrt(x^2+y^2) so you'd have sqrt(r^2).
 
sqrt r^2 is r right?
 
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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