What is the Correct Function for E in a Triple Integral?

[-EquinoX-]

Homework Statement

http://img5.imageshack.us/img5/5222/53026504.th.jpg

Homework Equations

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?

 

[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

 

[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?

 

[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.