By converting to polar coordinates, calculate

By converting to polar coordinates, calculate the following integral: $$ \iint_D xy\,dx\,dy, $$ where the region $D$ is bounded by the $x$-axis and the arcs of the circles $x^2+y^2=1$ and $x^2+y^2-2x=0$.

Step 1 · Analyze the circles

Circle 1: $x^2+y^2=1$ is centered at the origin with radius $1$; in polar form, $r=1$.

Circle 2: $x^2+y^2-2x=0 \Rightarrow (x-1)^2+y^2=1$, a circle centered at $(1,0)$ with radius $1$; in polar form, $r^2=2r\cos\theta \Rightarrow r=2\cos\theta$.

Step 2 · Find the region

The region $D$ lies above the $x$-axis, outside the unit circle, and inside the circle $r=2\cos\theta$. The circles intersect where $1=2\cos\theta$, i.e. $\cos\theta=\frac12$, giving $\theta=\frac{\pi}{3}$ (taking the branch above the $x$-axis). So for $0\leqslant\theta\leqslant\frac{\pi}{3}$, $r$ runs from $1$ to $2\cos\theta$.

Step 3 · Set up the integral

With $x=r\cos\theta$, $y=r\sin\theta$, $xy=r^2\cos\theta\sin\theta$, and $dx\,dy=r\,dr\,d\theta$:

$$ I = \int_0^{\pi/3}\int_1^{2\cos\theta} r^3\cos\theta\sin\theta\,dr\,d\theta. $$

Step 4 · Integrate with respect to $r$

$$ \int_1^{2\cos\theta} r^3\,dr = \left[\frac{r^4}{4}\right]_1^{2\cos\theta} = \frac{16\cos^4\theta-1}{4}, $$

so the inner integral is $\dfrac{\cos\theta\sin\theta}{4}\left(16\cos^4\theta-1\right)$.

Step 5 · Integrate with respect to $\theta$

$$ I = \frac14\int_0^{\pi/3}\left(16\cos^5\theta\sin\theta - \cos\theta\sin\theta\right)d\theta. $$

For the first term, let $u=\cos\theta$, $du=-\sin\theta\,d\theta$; as $\theta$ runs $0\to\frac\pi3$, $u$ runs $1\to\frac12$:

$$ \int_0^{\pi/3}16\cos^5\theta\sin\theta\,d\theta = 16\int_{1/2}^1 u^5\,du = 16\left[\frac{u^6}{6}\right]_{1/2}^1 = \frac{16}{6}\left(1-\frac{1}{64}\right) = \frac83\cdot\frac{63}{64}=\frac{21}{8}. $$

For the second term:

$$ \int_0^{\pi/3}\cos\theta\sin\theta\,d\theta = \frac12\int_0^{\pi/3}\sin2\theta\,d\theta = \left[-\frac{\cos2\theta}{4}\right]_0^{\pi/3} = -\frac14\left(-\frac12-1\right)=\frac38. $$

Step 6 · Combine the results

$$ I = \frac14\left(\frac{21}{8}-\frac38\right) = \frac14\cdot\frac{18}{8} = \frac{18}{32} = \frac{9}{16}. $$

Additional Notes

• Verified numerically by direct double integration in polar coordinates: $I\approx0.5625$, matching $\frac{9}{16}=0.5625$ exactly.