Area of Shapes - 1

We can do this using simple geometry; no calculus required. The area $A$ of the "missing piece" on top (bounded by the lines $y=\frac{1}{2}, x=\frac{1}{2}, x=-\frac{1}{2}$ and the circle) is the area of a sector of $60^o$ plus two triangles of area $\frac{1}{2}\frac{1}{2}\frac{\sqrt{3}}{2}$ each minus the area of a rectangle of area $1\times\frac{1}{2}$ . In summary, $A=\frac{\pi}{6}+\frac{\sqrt{3}}{4}-\frac{1}{2}$ . The area we seek is $\pi-4A=\frac{\pi}{3}-\sqrt{3}+2\approx \boxed{1.315}$ .

We know that the circle in total has an area of $\pi \cdot 1^2 = \pi$ . We notice that this shape misses only four parts of the circle. Moreover, those four parts can be split in 2 each and those are of equal size, because of the symmetry within the picture. One can notice that the area of one of those parts equals $\int_0^{\frac{1}{2}}\left(\sqrt{1-x^2}-\frac{1}{2}\right)dx$ To calculate this integral, we make use of the substitution $x = \sin(u)$ . It follows that $dx = \cos(u)du$ and thus, $\int_0^{\frac{1}{2}}\left(\sqrt{1-x^2}-\frac{1}{2}\right)dx = \int_0^{\frac{\pi}{6}}\left(\sqrt{1-\sin^2(u)} - \frac{1}{2}\right)\cdot \cos(u)du = \int_0^{\frac{\pi}{6}}\left(\cos^2(u) - \frac{\cos(u)}{2}\right)du$ Using the double angle formula for $\cos(2u)$ to our advantage, we get that $\cos^2(u)=\frac{\cos(2u)+1}{2}$ . Now it follows that $\left.\int_0^{\frac{\pi}{6}}\left(\cos^2(u) - \frac{\cos(u)}{2}\right)du = \int_0^{\frac{\pi}{6}}\left(\frac{\cos(2u)+1}{2} - \frac{\cos(u)}{2}\right)du = \frac{\sin(2u)}{4} - \frac{\sin(u)}{2} + \frac{u}{2}\right|_{u=0}^{u=\frac{\pi}{6}}$ Furthermore, we know that $\left.\frac{\sin(2u)}{4} - \frac{\sin(u)}{2} + \frac{u}{2}\right|_{u=0}^{u=\frac{\pi}{6}}=\frac{\sin(\frac{\pi}{3})}{4} - \frac{\sin(\frac{\pi}{6})}{2} + \frac{\pi}{12} - 0 = \frac{\sqrt{3}}{8}-\frac{1}{4}+\frac{\pi}{12}$ At last, we remember that this value equals the area of one out of eight missing parts of the shape to be a full circle with area $\pi$ . So, the area asked in the problem equals $\pi - 8 \cdot \left(\frac{\sqrt{3}}{8}-\frac{1}{4}+\frac{\pi}{12}\right) = \pi - \sqrt{3} + 2 - \frac{2\pi}{3} = \boxed{2+\frac{\pi}{3}-\sqrt{3}}$