# Is polar coordinate right?

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?

• Looks you took the opposite region, check once... – AgentS Mar 29 '15 at 11:06
• @ganeshie8 what do you mean by the opposite region? – mathsisfun Mar 29 '15 at 11:16
• you must get $0$ because the double integral refers to the x coordinate of center of mass of the given region. But it seems you took the opposite region, just double check your sketch... I think you should be working $$A = \int_{5\pi/6}^{13\pi/6}\int_{2sin\theta}^1 r \ cos\theta \ r \ dr \ d\theta$$ – AgentS Mar 29 '15 at 11:17
• @ganeshie8 the region D is the region above the circle centre(0,0) and r =1 and below the circle centre(0,1) r= 1. So should be from $\pi/6 to 5\pi/6$ right? – mathsisfun Mar 29 '15 at 11:23
• this green part is the region of integration right ? – AgentS Mar 29 '15 at 11:28

Hint: Convert the circle $x^2+(y-1)^2=1$ into polar coordinates: $r = 2 \sin \theta , 0 < \theta < \pi.$
• this what i did. then i $\int_{1}^{2sin\theta}$ – mathsisfun Mar 29 '15 at 11:27