The fact that the integral is proportional to the difference of the arithmetic and geometric means can be established in the following way, without calculating any integral. Let consider $\alpha=\frac{a+b}{2}$ and $\beta=\sqrt{ab}$ as independent variables. Then for the integral $$I(\alpha,\beta)=\int\limits_a^b \frac{\sqrt{(r-a)(b-r)}}{r}\,dr= \int\limits_a^b \frac{\sqrt{2r\alpha-r^2-\beta^2}}{r}\,dr,$$ we have $$\frac{\partial I}{\partial \alpha}=\int\limits_a^b \frac{dr}{\sqrt{2r\alpha-r^2-\beta^2}}=\int\limits_a^b \frac{dr}{\sqrt{(r-a)(b-r)}}$$ and $$\frac{\partial I}{\partial \beta}=-\int\limits_a^b \frac{\beta}{r\sqrt{2r\alpha-r^2-\beta^2}}\,dr=-\int\limits_a^b \frac{\sqrt{ab}}{r\sqrt{(r-a)(b-r)}}\,dr.$$ After substituting $$r=\frac{ab}{s}$$ in the last integral, we get $$\frac{\partial I}{\partial \beta}=-\int\limits_a^b \frac{\sqrt{ab}}{\sqrt{(ab-as)(bs-ab)}}\,ds=-\int\limits_a^b \frac{ds}{\sqrt{(b-s)(s-a)}}=-\frac{\partial I}{\partial \alpha} \tag{1}$$ Equation (1), combined with the fact, that for $\alpha=\beta=a$ (that is for $a=b$) the integral $I(a,a)=0$, implies that $$I(\alpha,\beta)=J(\alpha-\beta),$$ where $$J=\int\limits_a^b \frac{dr}{\sqrt{(r-a)(b-r)}}.$$

To complete the calculation, we need just to calculate $J$. This is done by the Euler substitution $$\sqrt{(r-a)(b-r)}=t(r-a)$$ which gives $$\int\limits_a^b \frac{dr}{\sqrt{(r-a)(b-r)}}=2\int\limits_0^\infty\frac{dt}{1+t^2}=\pi.$$

The fact that the integral is proportional to the difference of the arithmetic and geometric means can be established in the following way, without calculating any integral. Let consider $\alpha=\frac{a+b}{2}$ and $\beta=\sqrt{ab}$ as independent variables. Then for the integral $$I(\alpha,\beta)=\int\limits_a^b \frac{\sqrt{(r-a)(b-r)}}{r}\,dr= \int\limits_a^b \frac{\sqrt{2r\alpha-r^2-\beta^2}}{r}\,dr,$$ we have $$\frac{\partial I}{\partial \alpha}=\int\limits_a^b \frac{dr}{\sqrt{2r\alpha-r^2-\beta^2}}=\int\limits_a^b \frac{dr}{\sqrt{(r-a)(b-r)}}.$$ Note that this last integral does not depend on $a$ and $b$. Indeed, if we make a substitution $$x=\frac{r-a}{b-a},$$ we get $$\int\limits_a^b \frac{dr}{\sqrt{(r-a)(b-r)}}=\int\limits_0^1\frac{dx}{\sqrt{x(1-x)}}. $$ On the other hand, $$\frac{\partial I}{\partial \beta}=-\int\limits_a^b \frac{\beta}{r\sqrt{2r\alpha-r^2-\beta^2}}\,dr=-\int\limits_a^b \frac{\sqrt{ab}}{r\sqrt{(r-a)(b-r)}}\,dr.$$ After substituting $$r=\frac{ab}{s}$$ in the last integral, we get $$\frac{\partial I}{\partial \beta}=-\int\limits_a^b \frac{\sqrt{ab}}{\sqrt{(ab-as)(bs-ab)}}\,ds=-\int\limits_a^b \frac{ds}{\sqrt{(b-s)(s-a)}}=-\frac{\partial I}{\partial \alpha} \tag{1}$$ Equation (1), combined with the fact, that for $\alpha=\beta=a$ (that is for $a=b$) the integral $I(a,a)=0$, implies that $$I(\alpha,\beta)=J(\alpha-\beta),$$ where $$J=\int\limits_a^b \frac{dr}{\sqrt{(r-a)(b-r)}}=\int\limits_0^1\frac{dx}{\sqrt{x(1-x)}}.$$

To complete the calculation, we need just to calculate $J$. This is done by the Euler substitution $$\sqrt{(r-a)(b-r)}=t(r-a)$$ which gives $$J=\int\limits_a^b \frac{dr}{\sqrt{(r-a)(b-r)}}=2\int\limits_0^\infty\frac{dt}{1+t^2}=\pi.$$ Even a simpler option is to use the substitution $x=\sin^2\varphi$: $$J=\int\limits_0^1\frac{dx}{\sqrt{x(1-x)}}=2\int\limits_0^{\pi/2}d\varphi=\pi.$$