Circle Square Hybrid

We know that

$\displaystyle R_1=1$ and $\displaystyle R_2=\sec\theta\ \left\{-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\right\}$ denotes the right vertical edge of the square. Now,

$\displaystyle R_3=\frac{\left(1+\sec\theta\right)}{2}\left\{-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\right\}$ for the right bulging side of the third curve. Now, we simply calculate the area of this part and multiply it by $4$ since the figure is symmetrical.

Hence, we get area of one part

$\displaystyle \frac{1}{8}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(1+\sec\theta\right)^2d\theta$

$\displaystyle =\frac{1}{8}\left(\frac{\pi}{2}+4\ln\left(\sqrt{2}+1\right)+2\right)$

Hence, total area of the third curve is

$\displaystyle \frac{\pi}{4}+2\ln\left(\sqrt{2}+1\right)+1=3.54814....$

Similar solution with @Aaghaz Mahajan 's

We note that $R_1(\theta) = 1$ and $R_2(\theta) = \dfrac 1{\cos \theta} = \sec \theta$ . Then $R_3(\theta) = \dfrac {R_1(\theta) + R_2(\theta)}2 = \dfrac {1+\sec \theta}2$ , and the area enclosed by $R_3 (\theta)$ :

$\begin{aligned} A & = \int_0^{2\pi} \frac 12 R_3^2 (\theta) \ d\theta & \small \color{#3D99F6} \text{Since }R_3(\theta) \text{ is symmetrical about the origin} \\ & = 8 \int_0^\frac \pi 4 \frac 12 R_3^2 (\theta) \ d\theta \\ & = \int_0^\frac \pi 4 (1+\sec \theta)^2 \ d\theta \\ & = \int_0^\frac \pi 4 \left(1+ 2 \sec \theta + \sec^2 \theta \right) \ d\theta \\ & = \int_0^\frac \pi 4 d\theta + 2 \int_0^\frac \pi 4 \sec \theta\ d\theta + \int_0^\frac \pi 4 \sec^2 \theta \ d\theta \\ & = \left[\theta + 2 \int \frac {\tan \theta \sec \theta + \sec^2 \theta}{\tan \theta + \sec \theta } \ d\theta + \int d \tan \theta \right]_0^\frac \pi 4 \\ & = \bigg[\theta + 2 \ln (\tan \theta + \sec \theta) + \tan \theta \bigg]_0^\frac \pi 4 \\ & = \frac \pi 4 + 2 \ln (1+\sqrt 2) + 1 \\ & \approx 3.548145 \end{aligned}$

Therefore, $\lfloor 1000A \rfloor = \boxed{3548}$ .