Re: Integration in polar coordinates?
- From: big n <no@xxxxxxxxxxx>
- Date: Tue, 11 Apr 2006 11:59:20 -0400
On Mon, 10 Apr 2006 17:14:36 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@xxxxxxx> wrote:
Ah! I see. Thank you all for taking the time to answer my question!
If I may ask a follow-up -- what is the 3-D equivalent of this? Would
it just be r^2*dr*dt*dp?
Thanks again,
-n.
No, assuming you are talking about spherical coordinates (rho, theta,
phi) here, there is a sin(p) factor in addition to the r^2, which you
can verify with the Jacobian. Also, intuitively, if you are at (r, p,
t) and change each variable by dr, dp, and dt, respectively, the three
sides of the little "cube" are dr, r dp, and r sin(p) dt as the t
variable is in the xy plane and the radius for that is r sin(p).
Of course! I had not thought about the projection. Thank you for
your time.
-n.
.
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- Integration in polar coordinates?
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