How do I find a slice of the area of a semi-circle? A semi-circle around the origin, with radius $r$, is given by $$f(x) = \sqrt{r^2 - x^2}$$ The area of this semi-circle can be written as $$\int_{-r}^r \sqrt{r^2 - x^2} dx = \frac{\pi r^2}{2}$$
How do I find only a slice of this area, i.e. $$\int_a^b \sqrt{r^2 - x^2} dx$$ for $-r \le a &lt; b \le r$?

There's probably a geometric approach to this, but I'm kinda stuck. If it's easier to calculate a slice of the area of a full circle that's also fine.
Yes, this can be done in a geometric way. Assume that $a\leqslant 0\leqslant b$ and consider this picture:



The area that you are interested in is the area of two triangles plus the area of a circular sector. The areas of those triangles are$$\frac12(-a)\sqrt{r^2-a^2}\text{ and }\frac12b\sqrt{r^2-b^2}.$$and the area of the circular sector is$$\frac{r^2}2\left(\arccos\left(\frac ar\right)-\arccos\left(\frac br\right)\right).$$So, the area that you are interested in is the sum of these three numbers.

You will have to change a bit this argument in order to deal with the other two cases ($0\leqslant a\leqslant b\leqslant r$ and $-r\leqslant a\leqslant b\leqslant 0$).
Take $x = r\sin\theta$ so that $r^{2}-x^{2}$ becomes $r^{2}(1-\sin^{2}\theta) = r^{2}\cos^{2}\theta$ and $\sqrt{r^{2}-x^{2}}$ becomes $r\sin\theta$. Now, with this change of variable, one have $dx \to r\cos\theta d\theta$ and, thus:
$$\int_{a}^{b}\sqrt{r^{2}-x^{2}}dx = \int_{\sin^{-1}(a/r)}^{\sin^{-1}(b/r)}r^{2}\cos^{2}\theta d\theta  = r^{2}\bigg{(}\frac{1}{2}\theta+\frac{1}{4}\sin2\theta\bigg{)}\bigg{|}_{\sin^{-1}(a/r)}^{\sin^{-1}(b/r)}$$