# Why is the plot of bump function in two variables asymmetric?

I am interested in a function: $$\exp\left(-\frac{1}{1-x^2}\right)$$ on $(-1,1)$ and $0$ otherwise. According to Wikipedia this function is called bump function. In one variable it is a "bump". On the right side of the wikipedia page there is a plot of the bump function in two variables. It looks like croissant. I tried to plot $$\exp\left(-\frac{1}{1-x^2-y^2}\right)$$ in Wolfram alpha. It has circular symmetry. There is no croissant. What am I doing wrong?

• The Wikipedia picture illustrates a bump function, not "the bump function" that you have plotted... – Hans Lundmark Aug 30 '17 at 10:11

The bump function $f(x,y)$, defined on $x^2+y^2 \leq 1$ $$f(x,y) = \exp{\left(-\frac{1}{1-x^2-y^2}\right)}$$ and $f(x,y) = 0$ otherwise has radial symmetry.
To show it go in polar coordinates: \begin{align} x &= r \cos(\theta) \\ y &= r \sin(\theta) \end{align}
and observe that $r^2=x^2+y^2$ is the equation of a circle. So you get $$f(r) = \exp{\left(-\frac{1}{1-r^2}\right)}$$ so $f$ depends only by radius ($r$) and not angle.
The picture you refering to represents a bump function (it is smooth, compactly supported and not analytic) but it is not $f(x,y)$.