1.21. Compute the double integral $\iint_{\Omega} \sqrt{\left|x^{2}-y^{2}\right|} d x d y$, where
$\Omega=\{(x, z):|x|+|y| \leqslant a\}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-069.jpg?height=326&width=341&top_left_y=1072&top_left_x=174)

Fig. 1.42
Solution. To compute this integral, it is convenient to introduce new coordinates $u=x+y, v=x-y$. This transformation establishes a one-to-one correspondence between the square $\Omega:|x|+|y| \leqslant a$ in the $X O Y$ plane and the square $D:|u| \leqslant a,|v| \leqslant a$ in the UOV plane with the same origin $O$. The coordinate axes of these systems are rotated by $\frac{\pi}{4}$ relative to each other (Fig. 1.42). The Jacobian of this transformation $x=\frac{1}{2}(u+v), \quad y=\frac{1}{2}(u-v)$ is

$$
J=\frac{D(x, y)}{D(u, v)}=\left|\begin{array}{ll}
x_{u} & y_{u} \\
x_{v} & y_{v}
\end{array}\right|=\left|\begin{array}{cc}
\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2}
\end{array}\right|=-\frac{1}{2}
$$

The negative value of the Jacobian $J$ means that the systems $X O Y$ and $U O V$ have different orientations: the system $X O Y-$ is right-handed, and the system $U O V$ is left-handed. According to (1.36), the double integral will be transformed under the change of coordinates as follows:

$$
\begin{aligned}
& \iint_{\Omega} \sqrt{\left|x^{2}-y^{2}\right|} d x d y=\iint_{D} \sqrt{|u v|} \cdot|J| \cdot d u d v=\frac{1}{2} \int_{-a}^{+a} \sqrt{|u|} d u \int_{-a}^{+a} \sqrt{|v|} d v= \\
& \quad=2 \int_{0}^{a} \sqrt{u} d u \int_{0}^{a} \sqrt{v} d v=2\left(\frac{2}{3}\right)^{2} \cdot a^{3}=\frac{8}{9} a^{3} \cdot \\
& \text { 1.22. Compute the double integral } \iint_{\Omega} \frac{2 y}{x} e^{-\frac{\left(x^{2}+1\right) y}{x}} d x d y,
\end{aligned}
$$

where the region $\Omega$ is bounded by the lines

$$
x y=a>0, x y=b>0(a1
$$

Solution. Introduce new curvilinear coordinates $u, v$ by the formulas

$$
u=x y ; \quad v=\frac{y}{x}
$$

Then the region $\Omega$ on the $X O Y$ plane is transformed into the square $D=\left\{(u, v): a \leqslant u \leqslant b, \frac{1}{k} \leqslant v \leqslant k\right\}$ on the $u, v$ plane. Calculate the Jacobian of this transformation $x=\sqrt{\frac{u}{v}}, y=\sqrt{u v}$

$$
J=\frac{D(x, y)}{D(u, v)}=\left|\begin{array}{ll}
x_{u} & y_{u} \\
x_{v} & y_{v}
\end{array}\right|=\left|\begin{array}{cc}
\frac{1}{2} \frac{1}{\sqrt{u v}} & \frac{1}{2} \sqrt{\frac{v}{u}} \\
-\frac{1}{2} \sqrt{\frac{u}{v^{3}}} & \frac{1}{2} \sqrt{\frac{u}{v}}
\end{array}\right|=-\frac{1}{2 v}
$$

According to (1.36), the original double integral is transformed as follows:

$$
\begin{aligned}
& \iint_{\Omega} \frac{2 y}{x} e^{-\frac{\left(x^{2}+1\right) y}{x}} d x d y=\iint_{D} e^{-(u+y)} d u d v= \\
= & \int_{a}^{b} e^{-u} d u \int_{\frac{1}{k}}^{k} e^{-v} d v=\left(e^{-a}-e^{-b}\right)\left(e^{-1 / k}-e^{-k}\right)
\end{aligned}
$$

If $k \rightarrow \infty, a \rightarrow 0$, we obtain the value of the integral over the unbounded region $0 \leqslant x y \leqslant b, y \geqslant 0$ :

$$
\lim _{\substack{a \rightarrow 0 \\ k \rightarrow \infty}}\left(e^{-a}-e^{-b}\right)\left(e^{-1 / k}-e^{-k}\right)=1-e^{-b}
$$