Calculating integral with area of half circle I was recently told I can calculate this integral > $ \int_{-R_0}^{R_0} \frac{\sqrt{{R_0}^2-r^2}}{2} dr $ using the formula for area of half circle can someone please show this to me I cannot do it

This is basically the formula to get the area of a circle.

We define the function $f(r) = \sqrt{R_0^2 -r^2}$ for $r \in [-R_0, R_0]$. We see) that its graph is actually the semi "upper"-circle. (This is an easy application of Pythagoras theorem). This means, given some position $x$ on the x-axis, the height of this semi-circle will be $f(x)$.

Now let's think how we could compute the area of the half-circle : given a position $x$ on the $x$-axis, we can consider a small vertical rectangle of width $\Delta x$ that lies on the $x$-axis and of height $f(x)$ like this.

We then compute sum of the area of many small rectangles which will cover all the area of the semi-circle. Putting that in "integral term" and knowing that the area is actually $\pi R_0^2 /2$ we can conclude : $$ \int_{-R_0}^{R_0} f(r)\ dr = \pi \frac{R_0^2}{2} $$ Therefore your integral will be $\pi R_0^2 /4$.