Use polar coordinates to evaluate $$\iint_{D}^{} x \ dA$$where D is the region inside the circle, $x^2+(y-1)^2=1$ but outside the circle $x^2+y^2=1$
This what i have got so far: $$A = \int_{\pi/6}^{5\pi/6}\int_1^{2 \sin\theta} r \cos\theta \,r \,dr \, d\theta$$
upon integrating I'm getting $0$.
Is the area correct?
Hint: Convert the circle $ x^2+(y-1)^2=1 $ into polar coordinates: $ r = 2 \sin \theta , 0 < \theta < \pi. $