Double Integration in region $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.

Question

Calculate $$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy\,,$$ where $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.

I first found the intersection points that are $(1,0)$, $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{2}, \frac{1}{2}\right)$. And the determined the regions \begin{split} D & =\left\{(x,y)\in \mathbb{R}^{2}: x^{2}+y^{2}, y \le x\} = \{(r,\theta)\in \mathbb{R}^{2}: 0 \le r \le 1, 0 \le \theta \le \frac{\pi}{4} \right\}\\ B & =\left\{(x,y)\in \mathbb{R}^{2}: 0 \le x \le \frac{1}{\sqrt{2}}, 0 \le y \le x\right\}\\ C & =\left\{(x,y)\in \mathbb{R}^{2}: \frac{1}{\sqrt{2}} \le x \le 1, \frac{1}{\sqrt{2}} \le y \le 1-x \right\} \end{split} Then, $$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy = \iint_{D} (x^{2}+y^{2})^{-3/2} \,dx\,dy - \iint_{B} (x^{2}+y^{2})^{-3/2} \,dx\,dy - \iint_{C} (x^{2}+y^{2})^{-3/2} \,dx\,dy$$

Is this right? Or is there any other easiest way?

Answer 1

You are trying to integrate over the the shaded region in the diagram.

There are couple of ways to simplify this -

i) Use polar coordinates to avoid splitting the integral. In polar coordinates, $x = r \cos\theta, y = r \sin\theta, x^2+y^2 = r^2$

So line $x+y=1$ can be rewritten as $r ({\cos \theta + \sin\theta}) = 1$

leading to bounds $\frac{1}{\cos \theta + \sin\theta} \leq r \leq 1$

Also $y = x$ can be rewritten as $\tan\theta = 1, \theta = \frac{\pi}{4}$ leading to bounds $0 \leq \theta \leq \frac{\pi}{4}$

So the integral becomes,

$ \ \displaystyle \int_0^{\pi/4} \int_{1 / (\cos\theta + \sin\theta)}^1 \frac{1}{r^3} \cdot r \ dr \ d\theta = \int_0^{\pi/4} \int_{1 / (\cos\theta + \sin\theta)}^1 \frac{1}{r^3} \cdot r \ dr \ d\theta$

$\displaystyle = \int_0^{\pi/4} \big[-\frac{1}{r}\big]_{1 / (\cos\theta + \sin\theta)}^1 \ d\theta$

$\displaystyle = \int_0^{\pi/4} (\cos\theta + \sin\theta - 1) \ d\theta = 1 - \frac{\pi}{4} $

ii) Rotate the region by $\frac{\pi}{4}$ anti-clockwise (though not necessary here).

$x = r \cos (\theta - \frac{\pi}{4}), y = r \sin (\theta - \frac{\pi}{4}), x^2 + y^2 = r^2$

$x + y = 1$ becomes $\sqrt2 r \sin\theta = 1 \implies r = \frac{1}{\sqrt2 \sin\theta}$ and the integral becomes,

$\displaystyle \int_{\pi/4}^{\pi/2} \int_{\csc\theta / \sqrt2}^1 \frac{1}{r^2} \ dr \ d\theta $

Answer 2

In polar coordinates, the integration region is enclosed by $\theta \in (0,\frac\pi4)$ and $ r\in (\frac1{ \sin \theta + \cos\theta},1)$. Then, the integral becomes $$I=\int_0^{\pi4}\int_{\frac1{\sin \theta + \cos\theta}}^1 \frac1{r^2}drd\theta= \int_0^{\pi4}(\sin \theta +\cos \theta -1 )d\theta=1-\frac\pi4 $$