How to solve this integral with transformation to polar coordinates?

Question

How do I determine new limits when transforming to polar coordinates. I have this example, and I don't know how to solve it correctly. $$ \iint_D \frac{\ln\left(x^2+y^2\right)}{x^2+y^2}\,dx\,dy $$ where $D: 1\leq x^2+y^2\leq e^2.$

So I transformed $x$ and $y$ to polar coordinates: $x=r\cos\phi$ and $y=r\sin\phi$ ; $x^2+y^2=r^2\cos \phi +r^2\sin \phi=r^2.$

I got $$ \iint \frac{\ln\left(x^2+y^2\right)}{x^2+y^2}\,dx\,dy = \int d\phi \int \frac{\ln r^2}{r^2}r\,dr. $$ My question is how to determine the new limits of integration after transforming them to polar coordinates.

Answer 1

First off, $D$ is described in polar coordinates by $1\leq r\leq e$, and $0\leq \phi\leq 2\pi$. That's the easy part. You read the bounds for $r$ straight off the inequality for $x$ and $y$ (that inequality literally says $1\leq r^2\leq e^2$). And $D$ goes all the way around the origin, so $\phi$ goes from $0$ to $2\pi$. $D$ is rotationally symmetric, and bounded by circles centered at the origin, so this is as easy as it gets in polar coordinates.

The slightly more tricky part is that $dx\,dy$ does not just become $dr\,d\phi$. Just like in the one-dimensional case, there is an additional factor appearing here when doing substitution, related to the derivatives of one variable expressed as a function of the other.

In this case, $dx\,dy$ becomes $r\,dr\,d\phi$. So what you want is $$ \int_1^e\left(\int_0^{2\pi}\frac{\ln(r^2)}{r^2}r\,d\phi \right)dr\\ = 2\pi\int_1^e\frac{2\ln(r)}{r}\,dr $$ (where the inner $\phi$ integral disappears because the integrand does not depend on $\phi$).

Answer 2

Use the fact $\ln (r^2) = 2\ln r$. Then, integrate $\dfrac{\ln r}{r}$ which is easy as it is of the form $f'\times f$.

Edit Sorry I misunderstood your question. $D$ is the closed disk centered at $0$ with radius $e$, minus the open disk centered at $0$ with radius $1$, so in polar coordinates, $D = \left\{ re^{i\varphi} | 1 \leqslant r \leqslant e \text{ and } \varphi \in [0,2\pi) \right\}$

Answer 3

First draw your "D" in $xy$ plane. It is clearly annulus shaped. The region is basically between two circles of radius $1$ and $e$. This is your limits for $r$. To cover entire annulus part, we need $0\leq \phi \leq 2\pi$.