# Contour plot restricted to a region defined by multiple inequalities

I'm trying to use contour_plot for plotting contours defined only in a region of the parameter space, namely only in the triangle defined by the vertices $(0,0), (1/2,\sqrt{3}/2), (1,0)$. The faces of this triangle are defined by the intersections of the x-axis and the lines of equation $y=\sqrt{3}x$ and $y=-\sqrt{3}x+\sqrt{3}$.

Contour_plot has the nice option region, but it seems that region only accepts one condition, i.e. only one boundary because the documentation specifies that region must be of the form $f(x,y)$ and this is interpreted by the code of countour_plot as $f(x,y)>0$, i.e. countour_plot plots all points such that $f(x,y)>0$. Is there a way to do what I need with this version of contour_plot?

The following represents what I’d like to do but it does not work. It does not throw an error but only takes into account the first condition

contour_plot((x^2) * cos(x*y), (x,-10,5), (y,-5,5), region=(-sqrt(3)*x-y and -sqrt(3)*x+sqrt(3)-y), fill=True, contours=30)

Note that there are other imperfect solutions to this problem like using a white polygon to hide the part of the plot I do not want to see. However, this "solution" is not desirable because the maximum and minimum values of contour_plot will not be defined with respect to the region I am trying to show (I hope this is clear; I can show an example if this is not).

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Since both sqrt(3)*x-y and -sqrt(3)*x+sqrt(3)-y must be positive to define your region, multiply the two together:

var('x y')
contour_plot((x^2) * cos(x*y), (x,0,1), (y,0,sqrt(3)/2),
region=(sqrt(3)*x-y)*(-sqrt(3)*x+sqrt(3)-y), fill=True, contours=30)


Here's a live example.

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This solves the problem, thanks! Also, one might still want to use polygon to have better-looking boundaries.

( 2016-11-16 12:36:19 -0600 )edit

You can make the boundaries smoother by increasing plot points: plot_points=500

( 2016-11-16 13:20:38 -0600 )edit