The equation for determining the centre of mass (C.O.M) of a solid body are as follow:
$$x_{cm} = \frac{\int x dm}{\int dm}$$ $$y_{cm} = \frac{\int y dm}{\int dm}$$ $$z_{cm} = \frac{\int z dm}{\int dm}$$
where dm is the mass of a tiny element of the solid and x,y,z are the projections of the position vector of this element.
To calculate the COM of a semi-circular disc, we use the following procedure:
We are working in polar coordinates here since it helps us to exploit the symmetry of the problem.
Let $\vec{r}$ be the position vector of an element of the solid.
The x-projection of $\vec{r}$ is $\lvert \vec{r} \rvert \cos(\theta)$ where $\theta$ is a variable.
The mass density is defined as $\sigma = \frac{M}{A}$ where $M$ is the mass of the body and $A$ is the area of the body.
In our case, $\sigma$ is constant and equal to $\frac{m}{A}$. So $dm = \sigma dA$.
Plugging this into $\int x dm$, we get $\int \lvert \vec{r} \rvert \cos(\theta) \sigma dA$.
Now, the tutorial I was following changed the single integral to a double integral which was $\iint \lvert \vec{r} \rvert \cos(\theta) \sigma dA$.
This finally expands to $\int_0^R \int_0^{\pi} \lvert \vec{r} \rvert \cos(\theta) \sigma (r d\theta dr)$ which is $\int_0^R \int_0^{\pi} \sigma \lvert \vec{r} \rvert^2 \cos(\theta) d\theta dr$
I am aware of double integrals but I'm not sure whether that change from single to double integrals is completely legitimate and if it is, could I please know the name of it?