Double integral of $1/(x^2+y^2)$ restricted to $x^2+y^2\leq2$ and $x\leq1$

Question

Find $$\iint_D \frac{1}{(x^2+y^2)^2}dA$$ where $$D = \left\{ (x,y): x^2 + y^2 \leq 2 \right\} \cap \left\{ (x,y): x \geq 1 \right\}$$

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Because of the prevalence of $x^2+y^2$ terms here, I figured we would be using a change of variables to polar coordinates with $dA = rdrd\theta$. However, I ran into trouble when finding the bounds of the integral. I know $r$ goes from $0$ to $\sqrt 2$, but I got stumped when considering $\theta$. Solving for $\theta$ using the substitution $x=r\cos\theta$ into $x\leq 1$, I got $\theta =\arccos(\frac{1}{r})$. This seems ok, but it resulted in an integral that is impossible to solve by hand (maybe not technically impossible, but clearly I did something wrong here).

I also don't think using Cartesian coordinates would be the right approach, since the polar coordinate substitution results in very nice cancelling and easy integration.

Answer 1

Make the limits of $r$ a function of $\theta$.

$x = 1$ corresponds to $\theta = \pm \dfrac{\pi}{4} $

So for the interval $\theta \in [- \dfrac{\pi}{4}, \dfrac{\pi}{4} ] $ the limits for $r$ are $0$ and $\sec \theta$, otherwise, the limits are $0, \sqrt{2}$

Thus, the integral becomes

$$ I = \displaystyle \large\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^{\sec(\theta)} \dfrac{1}{r^2} r dr d\theta + \int_\frac{\pi}{4}^{\frac{7\pi}{4}} \int_1^\sqrt{2}\dfrac{1}{r^2} r dr d\theta $$

Answer 2

In polar coordinates, the area is enclosed by $$r=\sqrt2,\>\>\>\>\> r\cos\theta =1$$ Eliminate $r$ to get $\cos \theta =\frac1{\sqrt2}$, i.e. $\theta \in (-\frac\pi4, \frac\pi4)$. Therefore, the ares is integrated as $$A=\int_{-\frac\pi4}^{\frac\pi4}\int_\frac1{\cos \theta}^\sqrt2 \frac1{r^4}rdr d\theta=\frac14\int_{-\frac\pi4}^{\frac\pi4} \cos2\theta\>d\theta=\frac14 $$