Parabolic cylindrical coordinates

iNCREDiBLE
Messages
128
Reaction score
0
Can someone, please, show me an example of when you are better of with parabolic cylindrical coordinates than with cartesian coordinates when computing a triple integral over a solid?
 
Physics news on Phys.org
Things with symmetry around an axis, like a cylinder. Try this one with cartesian coordinates, then try it with cylindrical coordinates.

\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{\sqrt{x^2+y^2}}^{2} \,\,\,(x^2+y^2)\,\,dz\,dy\,dx
 
Last edited:
FrogPad said:
Things with symmetry around an axis, like a cylinder. Try this one with cartesian coordinates, then try it with cylindrical coordinates.

\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{\sqrt{x^2+y^2}}^{2} \,\,\,(x^2+y^2)\,\,dz\,dy\,dx

That works great in (circular) cylindrical coordinates, but not in parabolic cylindrical coordinates.
 
:blushing: hehe... sorry man.
I must have read it too quickly. Honestly, I've never worked in parabolic cylindrical coordinates.

have a good one :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top