How can I evaluate this double integral through a domain transformation?

$$\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\,\mathrm dx\mathrm dy$$ where $$D=\{(x,y)\in\mathbb{R}^2:\ 0\leq y\leq x\,, xy\leq 1\leq x+y-1\}$$

The region $D$ is the red one in this picture:

enter image description here

I have tried to convert $D$ to polar coordinates but then, it come up the following integral: $$\frac{1}{3}\int_0^{\frac{\pi}{4}}\cos 2t\left(\frac{1}{\sqrt{(\sin t\cos t)^3}}-\frac{8}{(\sin t+\cos t)^3}\right)dt$$ which I got stuck into.
Does someone any further ideas?

  • $\begingroup$ $$\int_0^{\pi/4}\int_{2/(\sin t+\cos t)}^{1/(\sin t\cos t)}r^2\cos2t\,\mathrm dr\mathrm dt$$ $\endgroup$ – Nosrati Aug 7 '18 at 13:27
  • $\begingroup$ The upper bound value is incorrect because $$xy\leq 1\ \Rightarrow\ r\cdot \cos t\cdot r\cdot \sin t\leq 1\ \Rightarrow\ r\leq\frac{1}{\sqrt{\cos t \sin t}}$$ $\endgroup$ – Samuel Leanza Aug 7 '18 at 17:15
  • $\begingroup$ You are right. . $\endgroup$ – Nosrati Aug 7 '18 at 17:18
  • $\begingroup$ Changing parameters to $u=x^2-y^2$ and $v=2xy$, as in $u+iv=(x+iy)^2$, helps in giving a very precise proof that the integral is not finite, but the same conclusion follows from the observation that for large $x$, and therefore small $y$, the function is not much smaller than $x$, which makes the integral over the area between $x$ and $x+\Delta x$ close to $\Delta x$. $\endgroup$ – random Aug 9 '18 at 14:24

I have found that is possible to solve this integral by the following transformation: $$\begin{cases} u=x+y\\ v=x-y\end{cases}\ \Rightarrow\ \begin{cases} x=\frac{u+v}{2}\\ y=\frac{u-v}{2}\end{cases}$$

The integral does not converge: $$\begin{eqnarray} \iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\ dxdy &=& \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{\frac{(u+v)^2}{4}-\frac{(u-v)^2}{4}}{\sqrt{\frac{(u+v)^2}{4}+\frac{(u-v)^2}{4}}}\cdot \left(-\frac{1}{2}\right)\ dv=\\ &=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{v}{\sqrt{u^2+v^2}}\ dv =\\ &=&-\frac{\sqrt{2}}{4} \int_2^{+\infty}u\ du\int_{\sqrt{u^2-4}}^u 2v(u^2+v^2)^{-1/2}\ dv=\\ &=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}u\cdot [\sqrt{u^2+v^2}]_{\sqrt{u^2-4}}^u\ du =\\ &=&-\int_2^{+\infty}u^2-u\sqrt{u^2-2}\ du=\\ &=&-\int_2^{+\infty}u^2\ du +\frac{1}{2}\int_2^{+\infty}2u(u^2-2)^{1/2}\ du=\\ &=&\lim\limits_{c\to +\infty}\left \{\left[-\frac{u^3}{3}\right]_2^c+\frac{1}{3}\left[\sqrt{(u^2-2)^3}\right]_2^c\right \}=\\ &=&\lim\limits_{c\to +\infty}\left(-\frac{c^3}{3}+\frac{8}{3}+\frac{1}{3}\sqrt{(c^2-2)^3}-\frac{2\sqrt{2}}{3}\right)=\\ &=&\frac{1}{3}\lim\limits_{c\to +\infty}(\sqrt{(c^2-2)^3}-c^3+8-2\sqrt{2})=-\infty\end{eqnarray}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.