Answer:
B
Step-by-step explanation:
It comes from straightforward application of the integral for path length.
[tex]\displaystyle S=\int{}\,ds=\int{\sqrt{dx^2+dy^2}}=\int_0^{2\pi}{\sqrt{(1-\sin(t))^2+(1-\cos(t))^2}}\,dt\\\\=\int_0^{2\pi}{\sqrt{(1-2\sin(t)+\sin^2(t))+(1-2\cos(t)+\cos(t)^2)}}\,dt\\\\=\int_0^{2\pi}{\sqrt{(2+\sin^2(t)+\cos^2(t))-2\sin(t)-2\cos(t)}}\,dt\\\\=\boxed{\int_0^{2\pi}{\sqrt{3-2\sin(t)-2\cos(t)}}\,dt}\qquad\text{matches B}[/tex]
