I have gotten this here integral to solve: $$\iint xy\,\mathrm dx\mathrm dy$$where the area is bounded by $x^2+y^2=1$, $x^2+y^2=2x$, $y=0$.
So, from the conditions we have: $$r^2=1,\ r=2\cos\theta,\ r\sin\theta =0.$$ If I draw a picture, I see that the area can be divided into two parts (above and below the x-axis), so I'd say it is sufficient to calculate the area of the part above the x-axis and multiply it by two to get the correct result.
Now, we're considering only the part above the x axis
From the last equation we can deduce that the angle goes from $ [0, \dfrac{\pi}{2}]$, since sine is zero at $\dfrac{\pi}{2}$.
Regarding the boundaries of $r$, I'm not sure how to determine them, because $r$ naturally is $[0,1]$. What I tried is $r \in [0,2\cos\theta]$
Now my integral is:
$$\int_0^{\frac{\pi}{2}}\int_0^{2\cos\theta} r^3cos\theta \sin\theta \,\mathrm dr\mathrm d\theta$$ which evaluates to $\dfrac{2}{3}$. After multiplying by $2$, I get $\dfrac{4}{3}$.
Is my solution correct?
There are multiple regions bound by $x^2+y^2 = 1$, $x^2+y^2 = 2x$ and $y = 0$. So the question should always state it clearly using inequality signs.
For example, if we are trying to integrate over the region shaded in the diagram, it should state
$x^2+y^2 \geq 1, x^2 + y^2 \leq 2x, y \geq 0$
Also note that at the intersection of both circles,
$x^2+y^2 = 2x = 1 \implies x = \cos\theta = \dfrac{1}{2}$. So $\theta = \dfrac{\pi}{3}$.
Therefore the integral for the region I have outlined in the diagram is,
$\displaystyle \int_0^{\pi/3} \int_1^{2\cos\theta} r^3 \sin\theta \cos\theta \ dr \ d\theta$