5 0 1 2 3 1 15 2 25 this uses v fz where fz

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ı 2πr ds, taken around the circle, giving 2π Ûc Ù ı 2πr (r dø) = c 0 So, the strength of the point source specified by F(z) = c lnz is just c. 2π (B) Combining Sources and Sinks Since the flux integral is zero away from any singularity, it follows that we can just add fields like the above to get an arbitrary configurations of sources and sinks with specified strengths ci by taking a sum of terms : c F(z) = £i i ln(z - zi) 2π -Ki lnz. Its imaginary part is given by the real part of the log, of 2π the magnitude, which tells us that the flow is circular. To see it exactly (and in which direction it goes) compute the velocity field: +Ki Kiz K“-y, x‘ v = F'(z) = = 2= 2 2 2πz– 2π|z| 2π(x + y ) which circulates counterclockwise if K is positive. To get the moment, we note that, since the circulation is in the xy-plane, it only has one coordinate: the z-coordinate. Therefore we need compute only one path integral (in the xy-plane). Now, actually we don't even need to evaluate the path integral, because of the following facts: (1) The complex path integral of 1/z around such a circle is equal to 2πi. (2) The imaginary part of the integral of 1/z (namely, 2π) equals the negative of t...
View Full Document

This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

Ask a homework question - tutors are online