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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?

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  • $\begingroup$ $$\int_0^{\pi/4}\int_{2/(\sin t+\cos t)}^{1/(\sin t\cos t)}r^2\cos2t\,\mathrm dr\mathrm dt$$ $\endgroup$ – 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
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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: $$\begin{eqnarray} \iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\ dxdy &=& \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{\frac{(u+v)^2}{4}-\frac{(u-v)^2}{4}}{\sqrt{\frac{(u+v)^2}{4}+\frac{(u-v)^2}{4}}}\cdot \left(-\frac{1}{2}\right)\ dv=\\ &=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{v}{\sqrt{u^2+v^2}}\ dv =\\ &=&-\frac{\sqrt{2}}{4} \int_2^{+\infty}u\ du\int_{\sqrt{u^2-4}}^u 2v(u^2+v^2)^{-1/2}\ dv=\\ &=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}u\cdot [\sqrt{u^2+v^2}]_{\sqrt{u^2-4}}^u\ du =\\ &=&-\int_2^{+\infty}u^2-u\sqrt{u^2-2}\ du=\\ &=&-\int_2^{+\infty}u^2\ du +\frac{1}{2}\int_2^{+\infty}2u(u^2-2)^{1/2}\ du=\\ &=&\lim\limits_{c\to +\infty}\left \{\left[-\frac{u^3}{3}\right]_2^c+\frac{1}{3}\left[\sqrt{(u^2-2)^3}\right]_2^c\right \}=\\ &=&\lim\limits_{c\to +\infty}\left(-\frac{c^3}{3}+\frac{8}{3}+\frac{1}{3}\sqrt{(c^2-2)^3}-\frac{2\sqrt{2}}{3}\right)=\\ &=&\frac{1}{3}\lim\limits_{c\to +\infty}(\sqrt{(c^2-2)^3}-c^3+8-2\sqrt{2})=-\infty\end{eqnarray}$$

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