# What is the correct way to calculate streamlines of a vector field

by Jonsson
Tags: correct, field, streamlines, vector
 P: 11 Hello there, What is wrong with my way of finding stream lines of a vector field? Say I have this vector field: $\vec{v} = x\,y\,\vec{i} + y\,\vec{j}$ You can see a plot here: http://kevinmehall.net/p/equationexp...0,10,-10,10%5D It appears as if the stream lines could be $y = log(x) + C$. I proceed to find out: $v_y \, \mathrm{d}x = v_x \, \mathrm{d}y\\ x\,y\,\mathrm{d}y = y\,\mathrm{d}x\\ \mathrm{d}y = \frac{1}{x}\,\mathrm{d}x\\ \int\,\mathrm{d}y = \int \frac{1}{x}\,\mathrm{d}x\\ y = log(x) + C$ This looks about right. However there is a problem, when I look back at my vector field (http://kevinmehall.net/p/equationexp...0,10,-10,10%5D), for values of x less than zero, it appears as if the streamlines should be a mirror-image of y = log(x) + C. So my question, does the above streamline calculation have more solutions which I have missed? Or is there something else which is wrong, which is causing me only to find the streamlines for x values greater than 0? Thank you for your time. Kind regards, Marius
 Mentor P: 9,730 Your derivation simply does not care about the orientation of the field. You could see y=0 as region where field lines from both sides end, as the field vanishes. The correct integral of ##\frac{1}{x}## is ##ln(|x|)+C## (where you can use a different C for positive and negative x), this allows to use negative x-values as well.
 P: 11 mfb, thank you so much :) Kind regards, Marius

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