1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Looks you took the opposite region, check once... $\endgroup$ – AgentS Mar 29 '15 at 11:06
  • $\begingroup$ @ganeshie8 what do you mean by the opposite region? $\endgroup$ – mathsisfun Mar 29 '15 at 11:16
  • $\begingroup$ 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 $$A = \int_{5\pi/6}^{13\pi/6}\int_{2sin\theta}^1 r \ cos\theta \ r \ dr \ d\theta$$ $\endgroup$ – AgentS Mar 29 '15 at 11:17
  • $\begingroup$ @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? $\endgroup$ – mathsisfun Mar 29 '15 at 11:23
  • $\begingroup$ this green part is the region of integration right ? $\endgroup$ – AgentS Mar 29 '15 at 11:28
0
$\begingroup$

Hint: Convert the circle $ x^2+(y-1)^2=1 $ into polar coordinates: $ r = 2 \sin \theta , 0 < \theta < \pi. $

$\endgroup$
  • $\begingroup$ this what i did. then i $\int_{1}^{2sin\theta}$ $\endgroup$ – mathsisfun Mar 29 '15 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.