Given a lamina $x^2 +y^2 = 1$ with density function $\sigma(x,y) = x+y$. Find the center of mass of the surface.
So first lets find the mass.
$$m= \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}(x+y)\, dx\, dy$$
We can convert it into polar coordinates
$$m= \int_{0}^{2\pi}\int_{0}^{1}(r \sin\theta +r\cos\theta)r \,dr \,d\theta$$
Solving it would result in $0$.
Why is that?
PS: I didn't write down the $M_x$ and $M_y$ here since with mass = $0$, I wouldn't be able to solve it anyway since I have to divide by $0$.
The unit disk $x^2 + y^2 \leq 1$ is symmetric under reflection through the origin. That is, for every point $(x,y)$ on the unit disk, the point $(-x,-y)$ also is on the unit disk.
Notice that according to your definitions, $$\sigma(-x,-y) = -x - y = -\sigma(x,y).$$ That is, for every point on the disk where the mass density is $\sigma(x,y)$, there is a corresponding point where the mass density is $-\sigma(x,y)$. The masses in the neighborhoods of these two points exactly cancel.
Is is unusual to have negative mass density at a point. I would check the setup of the problem to see if that is a mistake.