Double Integral $\iint\limits_D\frac{dx\,dy}{(x^2+y^2)^2}$ where $D=\{(x,y): x^2+y^2\le1,\space x+y\ge1\}$

Question

Let $D=\{(x,y)\in \Bbb R^2 : x^2+y^2\le1,\space x+y\ge1\}$. The integral to be calculated over $D$ is the following: \begin{equation} \iint_D \frac{dx\,dy}{(x^2+y^2)^2} \end{equation}

I do not know how to approach the problem. I have tried integrating the function in cartesian coordinates but it doesn't seem to work out. I have also tried the variable change $u=x^2+y^2$ and $v=x+y$ (with the associated jacobian transformation) and again I cannot obtain the result.

Answer 1

Among skills of a technical (as opposed to conceptual) nature that are used in calculus, among the ones in which most students are weakest is technical details of trigonometry. When changing this to polar coordinates, $x^2+y^2$ becomes $r^2$ and $dx\,dy$ becomes $r\,dr\,d\theta$ and the boundary at $x^2+y^2=1$ becomes $r=1$, but what is one to do with $x+y\ge 1$? Drawing the picture, you see that that's a diagonal line going through both the highest point on the circle (the one with the largest $y$-coordinate) and the rightmost point on the circle (the one with the largest $x$-coordinate). And the constraint says $r\cos\theta+r\sin\theta \ge 1$. So you want $\theta$ to go from $0$ to $\pi/2$ and $r$ to go from something to $1$. You can write $r\ge\dfrac 1 {\cos\theta+\sin\theta}$. Then you have $$ \int_0^{\pi/2} \int_{1/(\cos\theta+\sin\theta)}^1 \frac{r \, dr \,d\theta}{r^4} = \int_0^{\pi/2}\left( \int_{1/(\cos\theta+\sin\theta)}^1 \frac{dr}{r^3} \right) d\theta. $$ The inside integral is $$ \left.\frac{-1}{2r^2} \right|_{r:=1/(\cos\theta+\sin\theta)}^1 = \frac{-1} 2 + \frac 1 {2(\cos\theta+\sin\theta)^2}. $$ Then you need to integrate that from $0$ to $\pi/2$.

Alternatively, you could note that by circular symmetry, the integral is the same as $$ \iint_E \frac{dx\,dy}{(x^2+y^2)^2} \text{ where } E=\left\{(x,y): x^2+y^2 \le 1\ \&\ x\ge \frac{\sqrt 2} 2 \right\}. $$

Answer 2

Hint: try out with polar coordinates $x= r \cos(\theta) $, $y = r \sin(\theta) $, just be careful when "changing" the domain $D$.

Answer 3

Transforming to polar coordinates $(\rho,\phi)$, we find that on $x+y=1$ we have $\rho =1/(\cos(\phi)+\sin(\phi))$. Therefore, we can write

$$\begin{align} \int_D \frac{1}{(x^2+y^2)^2}\,dx\,dy&=\int_0^{\pi/2}\int_{1/(\sin(\phi)+\cos(\phi))}^1 \frac{1}{\rho^3}\,d\rho\,d\phi\\\\ &=\frac12\int_0^{\pi/2}\sin(2\phi)\,d\phi\\\\ &=\frac12 \end{align}$$