I am trying to integrate converting to polar coordinates, between two circles.
$$A = \iint_D x \,\mathrm{d}A $$
Ant the domain of integration is set to be the region in the first quadrant between the circles:
$x^2 + y^2 = 4$ and $x^2 + y^2 = 2x$
I have plotted the circles, and I suppose the region I am being asked for is what I have painted in red:
Now, I would know how to proceed if one circle was perfectly inside of the other (same center), I would set $\theta$ to be between $0$ and $\pi / 4$ and $r$ to be between the radius of the smallest circle and the radius of the other one. However, I cannot do any of those here.
So how should I convert this integral to a polar-coordinate integral?
Any help would be very appreciated.
Best Answer
For circle one, the equation is simply
$$C_1: r=2$$
For circle two $x^2+y^2=r^2,\quad x=r\cos\theta$, so
$$C_2: r^2=2r\cos\theta \implies r(r-2\cos\theta)=0 \implies r=0,r=2\cos\theta$$
Of these only the second solution is useful (the first just says $C_2$ passes through the origin), so the inner and outer radii for the integral are $2\cos\theta$ and $2$, respectively. And $\theta$ varies from $0$ to $\pi/2$ in the first quadrant.
Since $dA=r\,dr\,d\theta$, we get
$$A=\int_{0}^{\pi/2}{\int_{2\cos\theta}^{2}{r}\,dr\,d\theta}$$
(The original integral for area should have been $A=\iint{dA}$.)
If $I=\iint{x\,dA}$ is required, then from $x=r\cos\theta$
$$I=\int_{0}^{\pi/2}{\int_{2\cos\theta}^{2}{r^2\cos\theta}\,dr\,d\theta}$$