(after having spent some time working in the wrong region):
The statement of the boundaries is ambiguous, because they apply equally well to the blue or the green region shown in the graph above. But it wasn't a complete waste of time to integrate in the wrong place, as I will describe below.
The equation of the circle is $ \ x^2 \ + (y-1)^2 \ = \ 1 \ $ , so the functions for the "upper and lower" semicircles are $ \ y \ = \ 1 \ \pm \ \sqrt{1-x^2} \ . $ The blue region is bounded above by the upper semi-circle and below by the line $ \ x + y = 2 \ $ . The intersection points of the line and circle are $ \ (0,2) \ \text{and} \ (1,1) \ . $ So the double integral is
$$ \int_0^1 \int^{1 + \sqrt{1-x^2}}_{2-x} \ x \ \ dy \ dx \ = \ \int_0^1 \ xy \ \vert^{1 + \sqrt{1-x^2}}_{2-x} \ \ dx $$
$$ = \ \int_0^1 \ x \ \left[ \ (1 + \sqrt{1-x^2}) \ - \ (2-x) \ \right] \ \ dx \ = \ \int_0^1 \ x^2 \ - \ x \ + \ x \sqrt{1-x^2} \ \ dx $$
$$ = \ \left[ \ \frac{1}{3}x^3 \ - \ \frac{1}{2}x^2 \ - \ \frac{1}{3} (1-x^2)^{3/2} \ \right] \ \vert_0^1 $$
$$ = \ \left( \ \frac{1}{3} \ - \ \frac{1}{2} \ - \ 0 \ \right) \ - \left( \ 0 \ - \ 0 \ - \ \frac{1}{3} \ \right) \ = \ \frac{1}{6} \ , $$
which apparently does not agree with the "given answer". So who's right?
Let's take the integration over the green region, which is bounded below entirely by the lower semi-circle, bounded above by the upper semi-circle over $ \ [-1 , 0 ] \ $ , and above by the line over $ \ [0 , 1] \ . $ So we have two integrals here:
$$ \int_{-1}^0 \int^{1 + \sqrt{1-x^2}}_{1 - \sqrt{1-x^2}} \ x \ \ dy \ dx \ \ + \ \ \int_0^1 \int^{2-x}_{1 - \sqrt{1-x^2}} \ \ x \ \ dy \ dx $$
$$ = \ \ \int_{-1}^0 \ \ x \ \left[ \ (1 + \sqrt{1-x^2}) \ - \ ({1 - \sqrt{1-x^2}}) \ \right] \ \ dx $$
$$ + \ \ \int_0^1 \ x \ \left[ \ (2-x) \ - \ (1 - \sqrt{1-x^2}) \ \right] \ dx $$
$$ = \ \ 2 \ \int_{-1}^0 \ \ x \ \sqrt{1-x^2} \ \ dx \ + \ \ \int_0^1 \ x \ - \ x^2 \ + \ x \sqrt{1-x^2} \ \ dx $$
$$ = \ \left[ \ - \ \frac{2}{3} (1-x^2)^{3/2} \ \right] \ \vert_{-1}^0 \ + \ \ \left[ \ \frac{1}{2}x^2 \ - \ \frac{1}{3}x^3 \ - \ \frac{1}{3} (1-x^2)^{3/2} \ \right] \ \vert_0^1 $$
$$ = \ \left[ \ -\frac{2}{3} \ + \ 0 \ \right] \ + \ \left[ \ \left( \ \frac{1}{2} \ - \ \frac{1}{3} \ - \ 0 \ \right) \ - \left( \ 0 \ - \ 0 \ - \ \frac{1}{3} \ \right) \ \right] $$
$$ = \ -\frac{2}{3} \ + \ \frac{1}{6} \ + \ \frac{1}{3} \ = \ -\frac{1}{6} \ . $$
What was the point of doing this? Together, the green and blue regions cover the entire circle, which is symmetrical about the $ \ y- $ axis. We are integrating $ \ x \ $ , a function with odd symmetry (about the $ \ y-$ axis) over this region. So the integral of this function over the circle is zero. And indeed, the integrals over the blue and green regions have opposite signs.
So I believe the result of $ \ \frac{1}{6} \ $ for the blue region is reliable. My suspicion is that the solver may have made an error summing the fractions (I've seen it happen...).
Best Answer
Continue with
$$I= \iint_S (1-y) dS =\sqrt{3} \iint_{x>0,y>0,x+y<1}(1-y) dx dy\\ = \sqrt3 \int_0^1\int_0^{1-y} (1-y)dxdy = \sqrt3\int_0^1 (1-y)^2dy = \frac1{\sqrt3}$$