# Verification of a double integral with absolute value $\iint_{X} \frac{|xy|}{x^2+y^2} \ dxdy$

I wanted to evaluate the following integral,

$$I=\iint_{X} \frac{|xy|}{x^2+y^2} \ dxdy$$ with $${X}=\{(x,y): x^2+y^2\leq1, x^2+y^2\leq2|x|\}$$.

I used polar coordinates, setting $$x=r\cos(\theta), \ y=r\sin(\theta)$$, but I'm concerned about how to express the condition $$x^2+y^2\leq2|x|$$, which turns into $$r/2 \leq |cos(\theta)|$$, now since $$r\in[0,1]$$, it should be $$cos(\theta) \in [1/2, 1]$$, namely $$\theta\in[-\pi/3, \pi/3]\cup[2\pi/3, 4\pi/3]$$ so we end up evaluating 2 double integrals, the first one with theta in $$[-\pi/3, \pi/3]$$, the latter with theta in $$[2\pi/3, 4\pi/3]$$. Is my reasoning correct?

• Mikasa's answer is correct. Try to show that the area Mikasa has highlighted is $X$. Aug 11, 2022 at 12:53
• I'm not getting it. Look at the picture I posted, the intersection between the red-delimited area and the blue circles is not the one proposed by Mikasa Aug 11, 2022 at 13:40

In polar coordinates \begin{align} &\iint_X \frac{|xy|}{x^2+y^2}dxdy =4 \iint_{X,x>0, y>0}\frac{xy}{x^2+y^2}dxdy\\ =& \ 4\bigg( \int_{0}^{\pi/3} \int_0^1 +\int_{\pi/3}^{\pi/2}\int_0^{2\cos\theta}\bigg) \cos \theta \sin\theta \ rdr d\theta = \frac34 + \frac18=\frac78 \end{align}
It is enough to evaluate the integral in the region $$x,y\ge0$$, $$(x,y)$$ being an element in the intersection
• with the disk centered in $$(1,0)$$ of radius one, this is the right blue disk from the posted picture.
We further separate this region in two parts, separated by the vertical line of equation $$x=1/2$$. So let us compute: \begin{aligned} \frac I4 &= \iint_{\substack{(x,y)\in X\\ x,y\ge 0}} \frac{xy}{x^2+y^2}\; dx\; dy \\ &= \int_0^{1/2}dx \int_0^{\sqrt{2x-x^2}}\frac{xy}{x^2+y^2}\; dy + \int_{1/2}^1 dx \int_0^{\sqrt{1-x^2}}\frac{xy}{x^2+y^2}\; dy \\ &= \frac 12 \int_0^{1/2}dx \Big[x\log(x^2+y^2)\ \Big]_0^{\sqrt{2x-x^2}} + \int_{1/2}^1 dx \Big[x\log(x^2+y^2)\ \Big]_0^{\sqrt{1-x^2}} \\ &= \frac 12 \int_0^{1/2}x\Big(\log(x^2+(2x-x^2)) -\log x^2\Big)\; dx \\ &\qquad\qquad + \frac 12 \int_{1/2}^1 x\Big(\log(x^2+(1-x^2))-\log x^2\Big)\; dx \\ &= \frac 12 \int_0^{1/2}x\log(2x)\; dx - \frac 12 \int_0^1 x\log x^2\; dx \\ &= \frac 12 \left[ \frac 12 x^2\log(2x) - \frac14x^2 \right]_0^{1/2}x\; dx - \frac 12 \left[ \frac 12 x^2\log(x^2) - \frac12 x^2 \right]_0^1 \\ &= -\frac 1{32} + \frac 14=\frac{32}{8-1}=\frac7{32}\ . \\[2mm] &\qquad\text{ Alternatively, we can use Fubini the other way around:} \\[2mm] \frac I4 &= \iint_{\substack{(x,y)\in X\\ x,y\ge 0}} \frac{xy}{x^2+y^2}\; dx\; dy \\ &= \int_0^{\sqrt 3/2}dy \int_{1-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{xy}{x^2+y^2}\; dx \\ &= \frac12 \int_0^{\sqrt 3/2}y\; dy\; \Big[\ \log(x^2+y^2)\ \Big] _{1-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \\ &= \frac12 \int_0^{\sqrt 3/2}y\; dy\; \Big[\ \log\Big((1-y^2)+y^2\Big) - \log\Big(\Big(1-\sqrt{1-y^2}\Big)^2 + y^2\Big)\ \Big] \\ &= - \frac12 \int_0^{\sqrt 3/2}y\; dy\; \log\Big(\Big(1-\sqrt{1-y^2}\Big)^2+y^2\Big) \\ &= - \frac12 \int_0^{\sqrt 3/2}y\; dy\; \log\Big(2-2\sqrt{1-y^2}\Big) = - \frac14 \int_0^{3/4} \log\Big(2-2\sqrt{1-t}\Big)\; dt \\ &= - \frac14 \Big[\ t\log\Big(2-2\sqrt{1-t}\Big)\ \Big]_0^{3/4} + \frac14 \int_0^{3/4} t\cdot\frac 1{2-2\sqrt{1-t}}\cdot\frac 1{\sqrt{1-t}}\; dt =\dots \\ &=\frac7{32}\ . \end{aligned} At the last step use $$(2-2\sqrt{1-t}) (2+2\sqrt{1-t})=4-4(1-t)=4t$$. We obtain: $$I = \frac 78\ .$$