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I need to evaluate the following integral:

$$\int\int_G \frac{\ln{(x^2+y^2)}}{x^2+y^2} dx dy$$

Here $G=\{(x,y)\in \mathbb{R}^2: 1 \leq x^2+y^2 \leq e^2\}$. I tried to use parametric transformations $x=a \cos{\phi}$, $y = a\sin{\phi}$ where $a$ ranges from $1$ to $e$ and $\phi$ ranges from $0$ to $2\pi$. But I can't transform $dx$ and $dy$ into very usable forms with that. Any help? Thank you.

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2 Answers 2

up vote 3 down vote accepted

Whenever you transform coordinates, say $x = f(r, \theta)$ and $y = g(r,\theta)$, then $$dx dy = \vert J \vert dr d \theta$$ where $$J = \begin{bmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial \theta}\\ \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \theta}\end{bmatrix}$$ In your case, you will find that $\vert J \vert = r$ and hence $dx dy = r dr d\theta$. Hence, your integral becomes $$\int_1^e \int_0^{2 \pi} \dfrac{\ln(r^2)}{r^2} r dr d\theta = \int_1^e \int_0^{2 \pi} \dfrac{2\ln(r)}{r} dr d\theta = 4 \pi \int_1^e \dfrac{\ln(r)}r dr = 4 \pi \left.\dfrac{\log^2(r)}2 \right \vert_{1}^e = 2 \pi$$

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Change to polar coordinates: $x = r \cos{\theta}$, $y = \sin{\theta}$. Then an area element is $dx \, dy = r \, dr \, d \theta$ and the integral becomes

$$\begin{align} \int_0^{2 \pi} d \theta \; \int_1^e dr \: r \frac{\log{r^2}}{r^2} &= 2 \pi \frac{1}{2} \int_1^{e^2} dx \frac{\log{x}}{x} \\ &= \pi \frac{1}{2} [(\log{x})^2]_1^{e^2} \\ &= 2 \pi\end{align}$$

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