Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D \frac{\ln(x^2+y^2)}{x^2+y^2}\,dx\,dy, $$ where $D$ is the annulus $1\leqslant x^2+y^2\leqslant e^2$.
Explanation of Method
In polar coordinates the annulus becomes the simple rectangle $1\leqslant r\leqslant e$, $0\leqslant\theta\leqslant2\pi$, and the integrand $\frac{\ln(r^2)}{r^2}=\frac{2\ln r}{r^2}$ combines with the extra factor of $r$ from the area element to leave $\frac{2\ln r}{r}$, whose antiderivative is $(\ln r)^2$.
- Convert to polar coordinates.
- Integrate with respect to $r$ using $u=\ln r$.
- Integrate the constant result with respect to $\theta$.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, the annulus becomes $1\leqslant r\leqslant e$, $0\leqslant\theta\leqslant2\pi$, and $x^2+y^2=r^2$:
$$ I = \int_0^{2\pi}\int_1^e \frac{\ln(r^2)}{r^2}\cdot r\,dr\,d\theta = \int_0^{2\pi}\int_1^e \frac{2\ln r}{r}\,dr\,d\theta. $$
Step 2 · Integrate with respect to $r$
With $u=\ln r$, $du=\frac{dr}{r}$; when $r=1$, $u=0$, and when $r=e$, $u=1$:
$$ \int_1^e \frac{2\ln r}{r}\,dr = 2\int_0^1 u\,du = 2\cdot\frac12 = 1. $$
Step 3 · Integrate with respect to $\theta$
$$ I = \int_0^{2\pi} 1\,d\theta = 2\pi. $$
Additional Notes
- The inner radius $r=1$ makes $\ln r=0$ there, so the integrand vanishes at the inner boundary of the annulus, which is what keeps the integral finite despite the $\frac{1}{r^2}$ factor.
Result
The solution to the integral is: