Integrate $f(x, y) = \frac{\ln(x^2 + y^2)}{\sqrt{x^2+y^2}}$ over the region $1 \le x^2 + y^2 \le e^3$. Region not bounded with respect to $\theta$?

2.9k Views Asked by At

Integrate $f(x, y) = \dfrac{\ln(x^2 + y^2)}{\sqrt{x^2+y^2}}$ over the region $1 \le x^2 + y^2 \le e^3$.

We change our coordinates system to simplify this computation:

$\therefore 1 \le \rho^2\cos^2\theta + \rho\sin^2\theta \le e^3$

$\implies 1 \le \rho^2 \le e^3$

$\implies 1 \le \rho \le e^{3/2}$ since $\rho \ge 0$

So we can find the bounds of integration in terms of $\rho$. But it seems that it is not possible to find the bounds of integration with respect to $\theta$? Does this mean that the region is not bounded with respect to $\theta$? What is the correct way to conceptually understand/interpret this?

I would greatly appreciate it if people could please take the time to clarify this.

1

There are 1 best solutions below

3
On BEST ANSWER

The bounds of the integration are revealed by the geometry of the domain of integration. Your region is an annulus and so it is described in polar coordinates by

$$ 1 \leq \rho \leq e^{\frac{3}{2}}, \,\, 0 \leq \theta \leq 2\pi. $$

By changing the coordinates to polar coordinates, you are left with computing

$$ \int_0^{e^{\frac{3}{2}}} \int_0^{2\pi} \ln(\rho^2) \, d \rho \,d \theta. $$