Evaluate $\iint_D|x - y|\mathrm{d}x\mathrm{d}y$ where $D = \{(x,y)\in\mathbb{R}^{2} \mid x^{2} + y^{2} \leq 2(x + y)\}$

Question

I have problem with solving the following integral.

Given that $D = \{(x,y)| x^2 + y^2 \leq 2(x+y)\}$, evaluate the double integral:

$$\iint_D|x - y|\mathrm{d}x\mathrm{d}y$$

I think I should convert $D$ to polar coordinates using the formulas $x = r\cos(\theta)$ and $y = r\sin(\theta)$, but I can't find the range of $\theta$.

Thanks in advance!

Answer 1

Note that $D$ is a circle tangential to the line $y+x=1$ at the origin. In polar coordinates, $D = \{(r,\theta)| r< 2(\cos\theta+\sin\theta)\}$ with $\theta \in (-\frac\pi4, \frac{3\pi}4)$

\begin{align} &\iint_D|x - y|\mathrm{d}x\mathrm{d}y\\ =& \int_{-\frac\pi4}^{\frac{3\pi}4 }\int_0^{2(\cos\theta+\sin\theta)} r|\cos\theta - \sin\theta|\ rdrd\theta\\ =& \ 2\int_{-\frac\pi4}^{\frac{\pi}4 }\int_0^{2(\cos\theta+\sin\theta)}(\cos\theta - \sin\theta)r^2 dr d\theta\\ = & \ \frac{16}3\int_{-\frac\pi4}^{\frac{\pi}4 }(\cos\theta - \sin\theta)(\cos\theta+\sin\theta)^3d\theta=\frac{16}3 \end{align}

Answer 2

Hint: $x^2-2x+y^2-2x\leq 0$ is the disc $(x-1)^2+(y-1)^2\leq 2$ so first use a change of variables, $(x,y)=(u+1, v+1)$

$$\iint_{x^2+y^2\leq x+y} \lvert x-y\rvert\,\mathrm d (x,y) = \iint_{u^2+v^2\leq 2}\lvert u-v\rvert\,\mathrm d (u,v)$$

Then use $(u,v) =(r\cos\theta,r\sin\theta)$